This is Part 1 of our AP Calculus AB unit test on derivatives. These questions cover the limit definition of derivatives, basic derivative operations (power rule, product rule, quotient rule, chain rule), derivatives of trig functions, and derivatives of logarithmic and exponential functions. Knowing these derivative operations will be necessary when analyzing the behavior of functions, as well as solving application oriented problems on the AP exam.

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Question 1 |

**Evaluate the derivative of the following function using the limit definition:**$f(x) = x^2$

$x$ | |

$\text{DNE}$ | |

$2x$ | |

$2$ |

Question 1 Explanation:

The answer is (C). Recall that the limit definition is as follows.

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{f(x + ∆h) − f(x)}{∆h}$

After plugging the function into the limit definition, a problem can clearly be seen. The limit cannot be identified with $∆h$ being $0$, as it makes the limit undefined. Work must be done to try and simplify the expression so that the limit can be evaluated.

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{f(x + ∆h)^2 − x^2}{∆h}$

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{x^2 + 2x∆h + ∆h^2 − x^2}{∆h}$

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{2x∆h + ∆h^2}{∆h}$

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{∆h(2x + ∆h)}{∆h}$

$f'(x) =$ $\lim\limits_{∆h\to 0} (2x + ∆h) = 2x$

The derivative of $x^2$ using the limit definition is $2x$.

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{f(x + ∆h) − f(x)}{∆h}$

After plugging the function into the limit definition, a problem can clearly be seen. The limit cannot be identified with $∆h$ being $0$, as it makes the limit undefined. Work must be done to try and simplify the expression so that the limit can be evaluated.

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{f(x + ∆h)^2 − x^2}{∆h}$

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{x^2 + 2x∆h + ∆h^2 − x^2}{∆h}$

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{2x∆h + ∆h^2}{∆h}$

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{∆h(2x + ∆h)}{∆h}$

$f'(x) =$ $\lim\limits_{∆h\to 0} (2x + ∆h) = 2x$

The derivative of $x^2$ using the limit definition is $2x$.

Question 2 |

**Evaluate the derivative of the following function using the limit definition:**$f(x) = \cos(x)$

$f'(x) = \sin(x)$ | |

$f'(x) = -\sin(x)$ | |

$\text{DNE}$ | |

$f'(x) = 0$ |

Question 2 Explanation:

The correct answer is (B). Recall that the limit definition of the derivative is as follows.

$f'(x)$ $= \lim\limits_{∆h\to 0} \dfrac{f(x + ∆h) − f(x)}{∆h}$

Be sure to remember trig identities for limit definition problems.

$f'(x)$ $= \lim\limits_{∆h\to 0} \dfrac{\cos(x + ∆h) − \cos(x)}{∆h}$

$f'(x)$ $= \lim\limits_{∆h\to 0} \dfrac{\cos(x) \cos(∆h) − \sin(x) \sin (∆h) − \cos(x)}{∆h}$

Remember the cosine addition identity.

$\cos(a + b)$ $= \cos(a) \cos(b) − \sin(a) \sin(b)$

Then recall that limits can be distributed across terms that are added or subtracted.

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{\cos(x) (\cos(∆h) − 1)}{∆h} − \lim\limits_{∆h\to 0} \dfrac{\sin(x) \sin(∆h)}{∆h}$

It is necessary to recognize that the trigonometric functions of $x$ can be factored out of the distributed limits, because they are not dependent on the limit. In other words, because the $\cos(x)$ and $\sin(x)$ are not in terms of $∆h$, they can be factored out of the limit, as they are considered “constants” for the limit of $∆h$.

$f'(x) =$ $\cos (x) \lim\limits_{∆h\to 0} \dfrac{(\cos(∆h) − 1)}{∆h} − \sin(x) \lim\limits_{∆h\to 0} \dfrac{\sin(∆h)}{∆h}$

Finally, it is important to recognize two things that you should put to memory.

$\lim\limits_{θ\to 0} \dfrac{\cos(θ) − 1}{θ} = 0$

$\lim\limits_{θ\to 0} \dfrac{\sin(θ)}{θ} = 1$

Now, we can solve.

$f'(x) = \cos(x) (0) − \sin(x) (1)$ $= −\sin (x)$

$f'(x)$ $= \lim\limits_{∆h\to 0} \dfrac{f(x + ∆h) − f(x)}{∆h}$

Be sure to remember trig identities for limit definition problems.

$f'(x)$ $= \lim\limits_{∆h\to 0} \dfrac{\cos(x + ∆h) − \cos(x)}{∆h}$

$f'(x)$ $= \lim\limits_{∆h\to 0} \dfrac{\cos(x) \cos(∆h) − \sin(x) \sin (∆h) − \cos(x)}{∆h}$

Remember the cosine addition identity.

$\cos(a + b)$ $= \cos(a) \cos(b) − \sin(a) \sin(b)$

Then recall that limits can be distributed across terms that are added or subtracted.

$f'(x) =$ $\lim\limits_{∆h\to 0} \dfrac{\cos(x) (\cos(∆h) − 1)}{∆h} − \lim\limits_{∆h\to 0} \dfrac{\sin(x) \sin(∆h)}{∆h}$

It is necessary to recognize that the trigonometric functions of $x$ can be factored out of the distributed limits, because they are not dependent on the limit. In other words, because the $\cos(x)$ and $\sin(x)$ are not in terms of $∆h$, they can be factored out of the limit, as they are considered “constants” for the limit of $∆h$.

$f'(x) =$ $\cos (x) \lim\limits_{∆h\to 0} \dfrac{(\cos(∆h) − 1)}{∆h} − \sin(x) \lim\limits_{∆h\to 0} \dfrac{\sin(∆h)}{∆h}$

Finally, it is important to recognize two things that you should put to memory.

$\lim\limits_{θ\to 0} \dfrac{\cos(θ) − 1}{θ} = 0$

$\lim\limits_{θ\to 0} \dfrac{\sin(θ)}{θ} = 1$

Now, we can solve.

$f'(x) = \cos(x) (0) − \sin(x) (1)$ $= −\sin (x)$

Question 3 |

**Identify the derivative of the following:**$f(x) = 9x^4 + 3x$

$f'(x) = 36x + 3$ | |

$f'(x) = 36x^4 + 3$ | |

$f'(x) = 36x^3$ | |

$f'(x) = 36x^3 + 3$ |

Question 3 Explanation:

The correct answer is (D). This problem also requires the power rule. Be sure to remember the exponent change.

$f'(x) =$ $9(4)(x^{4 −1}) + 3(x^0)$ $= 36^3 + 3$

$f'(x) =$ $9(4)(x^{4 −1}) + 3(x^0)$ $= 36^3 + 3$

Question 4 |

**Identify the derivative of the following:**$f(x) = 6x \cdot \sqrt{x}$

$f(x) = 3x^{–\frac{1}{2}}$ | |

$f(x) = 9x^{\frac{1}{2}}$ | |

$f(x) = 3x^{\frac{1}{2}}$ | |

$f(x) = 9x^{−\frac{1}{2}}$ |

Question 4 Explanation:

The correct answer is (B).
This problem requires power rule and product rule. Be careful when solving.

$f'(x) =$ $6x(\frac{1}{2}(x)^{\frac{1}{2} − 1} + (\sqrt{x})(6)$ $= 3x (x^{−\frac{1}{2}}) + 6 (x^{\frac{1}{2}})$

$f'(x) = 9x^{\frac{1}{2}}$

You will likely notice that $f(x)$ can be simplified to $6x^{\frac{3}{2}}$ prior to taking the derivative. If you take the derivative from this point, you will still get the same answer.

$f'(x) =$ $6x(\frac{1}{2}(x)^{\frac{1}{2} − 1} + (\sqrt{x})(6)$ $= 3x (x^{−\frac{1}{2}}) + 6 (x^{\frac{1}{2}})$

$f'(x) = 9x^{\frac{1}{2}}$

You will likely notice that $f(x)$ can be simplified to $6x^{\frac{3}{2}}$ prior to taking the derivative. If you take the derivative from this point, you will still get the same answer.

Question 5 |

**Identify the derivative of the following:**$f(x)$ $= \dfrac{7x^3}{(x + 3)}$

$f'(x)$ $= \dfrac{14x^3 + 63x^2}{x^2 + 6x + 9}$ | |

$f'(x)$ $= \dfrac{14x^3 + 63x^2}{x + 3}$ | |

$f'(x)$ $= \dfrac{28x^3 + 63x^2}{x^2 + 6x + 9}$ | |

$f'(x)$ $= \dfrac{21x^3 + 63x^2}{(x + 3)^2}$ |

Question 5 Explanation:

The correct answer is (A). Quotient rule is necessary to solve this problem.

$f'(x) =$ $\dfrac{(x + 3)(21x^2) − (7x^3)(1)}{(x + 3)^2}$

$f'(x) =$ $\dfrac{21x^3 + 63x^2 − 7x^3}{(x + 3)^2}$ $= \dfrac{14x^3 + 63x^2}{x^2 + 6x + 9}$

No further simplification can be done, so this is the final answer. You may notice that this problem can also be done using product rule when $f(x)$ $= 7x^3(x + 3)^{-1}$

$f'(x) =$ $\dfrac{(x + 3)(21x^2) − (7x^3)(1)}{(x + 3)^2}$

$f'(x) =$ $\dfrac{21x^3 + 63x^2 − 7x^3}{(x + 3)^2}$ $= \dfrac{14x^3 + 63x^2}{x^2 + 6x + 9}$

No further simplification can be done, so this is the final answer. You may notice that this problem can also be done using product rule when $f(x)$ $= 7x^3(x + 3)^{-1}$

Question 6 |

**Identify the derivative of the following:**$f(x) = 7\cos(3x) \cdot 2\sin(2x)$

$f'(x) =$ $–14\sin(3x) \cdot \sin(2x)$ $+ 14\cos(3x) \cdot \cos(2x)$ | |

$f'(x) =$ $–42\cos(3x) \cdot \sin(2x)$ $+ 28\cos(2x) \cdot \sin(3x)$ | |

$f'(x) =$ $–42\sin(3x) \cdot \sin(2x)$ $+ 28\cos(3x) \cdot \cos(2x)$ | |

$f'(x) =$ $42\sin(3x) \cdot \sin(2x)$ $+ 28\cos(3x) \cdot \cos(2x)$ |

Question 6 Explanation:

The correct answer is (C). All the rules of basic derivatives still apply to trig functions, making product rule necessary for this problem. Don’t forget trig derivatives, as:

$\frac{d}{dx}(\sin(x))$ $= \cos(x)$ and $\frac{d}{dx}(\cos(x))$ $= −\sin(x)$

$f'(x) =$

$7(–\sin(3x))(3)(2\sin(2x))$ $+ (7\cos(3x))(2\cos(2x)(2))$

$f'(x) =$ $−42\sin(3x) \cdot \sin(2x)$ $+ 28\cos(3x) \cdot \cos(2x)$

$\frac{d}{dx}(\sin(x))$ $= \cos(x)$ and $\frac{d}{dx}(\cos(x))$ $= −\sin(x)$

$f'(x) =$

$7(–\sin(3x))(3)(2\sin(2x))$ $+ (7\cos(3x))(2\cos(2x)(2))$

$f'(x) =$ $−42\sin(3x) \cdot \sin(2x)$ $+ 28\cos(3x) \cdot \cos(2x)$

Question 7 |

**Identify the derivative of the following:**$f(x) = \cos^2(4x^2)$

$f'(x) =$ $–2\cos(4x^2)\sin(x)$ | |

$f'(x) =$ $–2\cos(4x^2)\sin(4x^2)$ | |

$f'(x) =$ $16x \cdot \cos(4x^2)\sin(4x^2)$ | |

$f'(x) =$ $–16x \cdot \cos(4x^2)\sin(4x^2)$ |

Question 7 Explanation:

The correct answer is (D). Power rule and chain rule must be used to solve this problem.

It may be simpler to recognize how to derive this function when it is rewritten in the form below, as the necessity for power rule may be more clearly seen in this form.

$f(x) =$ $(\cos(4x^2))^2$

Then, using power rule and chain rule,

$f'(x) =$ $2(\cos(4x^2))^{2 − 1} (−\sin(4x^2)(8x))$

$f'(x) =$ $−16x\cos(4x^2) \cdot \sin(4x^2)$

It may help to replace certain parts of difficult functions with a different variable. For example, this problem can be done by replacing $\cos(4x^2)$ with $u$, where the function is now of the form below.

$f(x) = u^2$

$f'(x) = 2(u)^1(u')$

After replacing the variable $u$ with the corresponding values, we get the same answer.

$f'(x) =$ $2(\cos(4x^2)) (−\sin(4x^2)(8x))$

$f'(x) =$ $−16(\cos(4x^2)) \cdot \sin(4x^2)$

Be careful with your derivation. Be sure to include all the parts.

It may be simpler to recognize how to derive this function when it is rewritten in the form below, as the necessity for power rule may be more clearly seen in this form.

$f(x) =$ $(\cos(4x^2))^2$

Then, using power rule and chain rule,

$f'(x) =$ $2(\cos(4x^2))^{2 − 1} (−\sin(4x^2)(8x))$

$f'(x) =$ $−16x\cos(4x^2) \cdot \sin(4x^2)$

It may help to replace certain parts of difficult functions with a different variable. For example, this problem can be done by replacing $\cos(4x^2)$ with $u$, where the function is now of the form below.

$f(x) = u^2$

$f'(x) = 2(u)^1(u')$

After replacing the variable $u$ with the corresponding values, we get the same answer.

$f'(x) =$ $2(\cos(4x^2)) (−\sin(4x^2)(8x))$

$f'(x) =$ $−16(\cos(4x^2)) \cdot \sin(4x^2)$

Be careful with your derivation. Be sure to include all the parts.

Question 8 |

**Identify the derivative of the following:**$f'(x) = \tan(\sec(x))$

$f'(x) =$ $\tan^2(\sec(x)) \cdot \sec(x) \tan(x)$ | |

$f'(x) =$ $\sec^2(\sec(x)) \cdot \sec(x) \tan(x)$ | |

$f'(x) =$ $\sec^2(x) \cdot \sec(x) \tan(x)$ | |

$f'(x) =$ $\sec^2(\sec(x)) \cdot \sec^2(x)$ |

Question 8 Explanation:

The correct answer is (B). Recall that the derivative of $\tan(u)$ is: $\sec^2(u) \cdot du$

Then, in the given problem, the “$u$” is $\sec(x)$. The derivative of $u$ is then $\frac{d}{dx}(\sec(x)) = \sec(x) \tan(x)$. Then, by chain rule:

$f'(x) =$ $\sec^2(\sec(x)) \cdot \sec(x) \tan(x)$

Then, in the given problem, the “$u$” is $\sec(x)$. The derivative of $u$ is then $\frac{d}{dx}(\sec(x)) = \sec(x) \tan(x)$. Then, by chain rule:

$f'(x) =$ $\sec^2(\sec(x)) \cdot \sec(x) \tan(x)$

Question 9 |

**Identify the derivative of the following:**$f(x) = \cot^{–1}(3x) + 2 \csc^{–1}(x^3)$

$f'(x) =$ $\dfrac{1}{1 + 9x^2} + \dfrac{2}{|x^3|\sqrt{x^6 − 1}}$ | |

$f'(x) =$ $\dfrac{–1}{1 + 9x^2} − \dfrac{2}{|x^3|\sqrt{x^6 − 1}}$ | |

$f'(x) =$ $\dfrac{3}{1 + 9x^2} + \dfrac{6x^2}{|x^3|\sqrt{x^6 − 1}}$ | |

$f'(x) =$ $\dfrac{–3}{1 + 9x^2} − \dfrac{6x^2}{|x^3|\sqrt{x^6 − 1}}$ |

Question 9 Explanation:

The correct answer is (D). Be sure to remember chain rule, as all regular derivative rules still apply when deriving inverse trig functions. Also, recall that:

$\dfrac{d}{du}(\cot^{–1}(u)) =$ $−\dfrac{1}{1 + u^2} \cdot du$

and

$\dfrac{d}{du}(\csc^{−1}(u)) =$ $−\dfrac{1}{|u|\sqrt{u^2 − 1}} \cdot du$

Then, we can substitute for u in the given function to arrive at the derivative below:

$f'(x) =$ $\dfrac{–1}{1 + (3x)^2} \cdot 3$ $− \dfrac{2}{|x^3|\sqrt{x^6 − 1}} \cdot 3x^2$

After simplifying, we arrive at the result below. Be careful not to simplify the absolute value:

$f'(x) =$ $\dfrac{–3}{1 + 9x^2} $ $− \dfrac{6x^2}{|x^3|\sqrt{x^6 − 1}}$

$\dfrac{d}{du}(\cot^{–1}(u)) =$ $−\dfrac{1}{1 + u^2} \cdot du$

and

$\dfrac{d}{du}(\csc^{−1}(u)) =$ $−\dfrac{1}{|u|\sqrt{u^2 − 1}} \cdot du$

Then, we can substitute for u in the given function to arrive at the derivative below:

$f'(x) =$ $\dfrac{–1}{1 + (3x)^2} \cdot 3$ $− \dfrac{2}{|x^3|\sqrt{x^6 − 1}} \cdot 3x^2$

After simplifying, we arrive at the result below. Be careful not to simplify the absolute value:

$f'(x) =$ $\dfrac{–3}{1 + 9x^2} $ $− \dfrac{6x^2}{|x^3|\sqrt{x^6 − 1}}$

Question 10 |

**Identify the derivative of the following:**$f(x) = \sec^{–1}(\sin(x) + 6)$

$f'(x) =$ $\dfrac{\cos(x)}{|\sin(x) +6|\sqrt{((\sin(x) + 6)^2) − 1}}$ | |

$f'(x) =$ $\dfrac{1}{|\sin(x) +6|\sqrt{((\sin(x) + 6)^2) − 1}}$ | |

$f'(x) =$ $− \dfrac{\cos(x)}{|\sin(x) +6|\sqrt{((\sin(x) + 6)^2) −1}}$ | |

$f'(x) =$ $\sec(\sin(x) + 6) \cdot \tan(\sin(x) +6) \cdot \cos(x)$ |

Question 10 Explanation:

The correct answer is (A). This may seem difficult, but it can be done with careful work. Be careful with chain rule. Also be sure to recognize the difference between the derivative of a traditional trigonometric function and an inverse trigonometric function. Recall that:

$\dfrac{d}{du}(\sec^{–1}(u))$ $= \dfrac{1}{|u|\sqrt{u^2 − 1}} \cdot du$

$f'(x) =$ $\dfrac{1}{|\sin(x) +6|\sqrt{((\sin(x) + 6)^2) − 1}} \cdot \cos(x)$

After further Simplifying, we arrive at the result below:

$f'(x) =$ $\dfrac{\cos(x)}{|\sin(x) +6|\sqrt{((\sin(x) + 6)^2) − 1}}$

$\dfrac{d}{du}(\sec^{–1}(u))$ $= \dfrac{1}{|u|\sqrt{u^2 − 1}} \cdot du$

$f'(x) =$ $\dfrac{1}{|\sin(x) +6|\sqrt{((\sin(x) + 6)^2) − 1}} \cdot \cos(x)$

After further Simplifying, we arrive at the result below:

$f'(x) =$ $\dfrac{\cos(x)}{|\sin(x) +6|\sqrt{((\sin(x) + 6)^2) − 1}}$

Question 11 |

**Identify the derivative of the following:**$f(x) = \ln(3x)$

$f'(x) = \frac{3}{x}$ | |

$f'(x) = \frac{1}{3x}$ | |

$f'(x) = \frac{1}{x}$ | |

$f'(x) = 3 \cdot \ln(3x)$ |

Question 11 Explanation:

The correct answer is (C). The use of $u$ replacements may be very useful for natural function derivation.

$u = 3x$

$f(x) = \ln(u)$

$f'(x) = \dfrac{1}{u}(u')$ $= \dfrac{1}{3x}(3)$ $= \dfrac{1}{x}$

This can be done without the use of $u$, but this tool may make more complicated problems easier in the future.

$u = 3x$

$f(x) = \ln(u)$

$f'(x) = \dfrac{1}{u}(u')$ $= \dfrac{1}{3x}(3)$ $= \dfrac{1}{x}$

This can be done without the use of $u$, but this tool may make more complicated problems easier in the future.

Question 12 |

**Evaluate the derivative of the following:**$f(x) = 7e^{3x +7}$

$f'(x) =$ $7e^{3x +7}$ | |

$f'(x) =$ $21x + 49 + 7e^{3x +7}$ | |

$f'(x) =$ $21xe^{3x +7} + 49e^{3x +7}$ | |

$f'(x) =$ $21e^{3x +7}$ |

Question 12 Explanation:

The correct answer is (D). It may be useful to replace the exponent with a variable $u$. Recall that:

$\dfrac{d}{du}(e^u) $ $= e^u \cdot du$

No matter how it is solved, the answer is of the form below:

$f'(x) =$ $7(e^{3x +7}(3)) $ $= 21e^{3x +7}$

All rules of basic derivation apply to derivatives of natural functions. Be sure to use the rules when they apply.

$\dfrac{d}{du}(e^u) $ $= e^u \cdot du$

No matter how it is solved, the answer is of the form below:

$f'(x) =$ $7(e^{3x +7}(3)) $ $= 21e^{3x +7}$

All rules of basic derivation apply to derivatives of natural functions. Be sure to use the rules when they apply.

Question 13 |

**Identify the derivative of the following:**$f(x) = 10^x$

$f'(x) = \ln(10) \cdot 10^x$ | |

$f'(x) = 10^x$ | |

$f'(x) = \ln(10)$ | |

$f'(x) = \ln(10)^x$ |

Question 13 Explanation:

The correct answer is (A). Unfortunately, basic derivative rules do not work for exponentials with a base other than $e$. The natural log can be used in exponential problems to make deriving simpler:

$\ln(f(x)) $ $ = \ln(10)^x $ $ = x \cdot \ln(10)$

Now $\ln(10)$ is just a constant, making deriving rather simple. However, it is important to recognize that you are no longer just deriving $f(x)$. Be sure to take the derivative of both sides of the equation accurately:

$\dfrac{1}{f(x)} \cdot f'(x)$ $= \ln(10)$

It is important to isolate $f'(x)$ when using the natural log tool:

$f'(x) = \ln(10) \cdot f(x) $ $= \ln(10) \cdot 10^x$

Be sure to replace $f(x)$ with the original function.

$\ln(f(x)) $ $ = \ln(10)^x $ $ = x \cdot \ln(10)$

Now $\ln(10)$ is just a constant, making deriving rather simple. However, it is important to recognize that you are no longer just deriving $f(x)$. Be sure to take the derivative of both sides of the equation accurately:

$\dfrac{1}{f(x)} \cdot f'(x)$ $= \ln(10)$

It is important to isolate $f'(x)$ when using the natural log tool:

$f'(x) = \ln(10) \cdot f(x) $ $= \ln(10) \cdot 10^x$

Be sure to replace $f(x)$ with the original function.

Question 14 |

**Identify the derivative of the following:**$f(x) = \log_3(\tan(2x))$

$f'(x) =$ $\dfrac {2\sec^2(2x)}{\ln(3)\tan(2x)}$ | |

$f'(x) =$ $\dfrac {1}{\ln(3)\tan(2x)}$ | |

$f'(x) =$ $\dfrac {2\sec^2(2x)}{\tan(2x)}$ | |

$f'(x) =$ $\dfrac {1}{\tan(2x)}$ |

Question 14 Explanation:

The correct answer is (A). Unfortunately, the derivation rules for the natural logarithmic function do not apply to logarithms of bases that are not $e$. Therefore, we must first convert the given function into a logarithm with base $e$.

$f(x) =$ $\log_3(\tan(2x))$

$= \dfrac{\ln(\tan(2x))}{\ln(3)}$

Then, we can now take the derivative. It may be beneficial to recognize that the denominator can be factored out as a coefficient, in order to avoid the use of the laborious quotient rule. Also, recall that: $\frac{d}{du}(\ln(u)) = \frac{1}{u} \cdot du$

$f'(x) =$ $\dfrac{1}{\ln(3)} \cdot \dfrac{1}{\tan(2x)} \cdot 2\sec^2(2x)$

$= \dfrac {2\sec^2(2x)}{\ln(3)\tan(2x)}$

$f(x) =$ $\log_3(\tan(2x))$

$= \dfrac{\ln(\tan(2x))}{\ln(3)}$

Then, we can now take the derivative. It may be beneficial to recognize that the denominator can be factored out as a coefficient, in order to avoid the use of the laborious quotient rule. Also, recall that: $\frac{d}{du}(\ln(u)) = \frac{1}{u} \cdot du$

$f'(x) =$ $\dfrac{1}{\ln(3)} \cdot \dfrac{1}{\tan(2x)} \cdot 2\sec^2(2x)$

$= \dfrac {2\sec^2(2x)}{\ln(3)\tan(2x)}$

Question 15 |

**Evaluate the derivative at $x = 3$ for the following:**$f(x) = 2e^x \cdot \sin (πx)$

$f'(3) = 0$ | |

$f'(3) = 2πe^3$ | |

$f'(3) = −2πe^3$ | |

$f'(3) = −2e^3$ |

Question 15 Explanation:

The correct answer is (C). As the function is a product of two functions, we must use product rule when taking the derivative. Don’t forget derivation rules for trigonometric and exponential functions.

$f'(x) =$ $2e^x(π \cos(πx)) + 2\sin(πx)(e^x)$

$= 2e^x(π \cos(πx) + \sin(πx))$

Then, don’t forget to evaluate the derivative at the given value, $x = 3$. Therefore:

$f'(3) =$ $2e^3(π \cos(3π)) + 2\sin(3π)$

$= 2e^3(π (−1) + 0)$

$= −2πe^3$

$f'(x) =$ $2e^x(π \cos(πx)) + 2\sin(πx)(e^x)$

$= 2e^x(π \cos(πx) + \sin(πx))$

Then, don’t forget to evaluate the derivative at the given value, $x = 3$. Therefore:

$f'(3) =$ $2e^3(π \cos(3π)) + 2\sin(3π)$

$= 2e^3(π (−1) + 0)$

$= −2πe^3$

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