MATH 1020 Chain Rule Learning Activity 4.3C 4. ( 6 pts ) Find the derivative of each of the following functions. $$ S(x) $$ You need not take the time to simplify your derivatives. a. $$ f(x)=4\left(3 x^{2}+2^{-6 x}\right)^{10} $$ $$ \begin{array}{l} f^{\prime}(x)=4 \cdot 10(5(x))^{9} \cdot 5^{\prime}(x)=4\left(3 x^{2}+2^{-6 x}\right)^{9} 5^{\prime}(x) \ f^{\prime}(x)=40\left(3 x^{2}+2^{-6 x}\right)^{9}\left(6 x-6 \ln (2) 2^{-6 x}\right) \end{array} $$ b. $$ g(x)=15 \ln \left(-3 e^{\frac{2}{3 x}}\right) $$ c. $$ h(t)=\frac{1}{\sqrt[3]{(3 t-4)^{5}}}+\ln \left(t^{4}\right) $$ d. $$ y=\frac{3}{5\left(e^{12 x+5}-6\right)^{3}}+\pi^{2} $$ e. $$ f(m)=\frac{1}{\sqrt{12 \ln (3 m)}} $$ f. $$ g(x)=\left(7^{\ln (x)}\right)^{2} $$

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a. We have 
$$
f(x)=4\Bigl(3x^2+2^{-6x}\Bigr)^{10}.
$$
Apply the chain rule. Taking the derivative we obtain
$$
f'(x)=4\cdot 10\Bigl(3x^2+2^{-6x}\Bigr)^{9}\cdot\frac{d}{dx}\Bigl(3x^2+2^{-6x}\Bigr).
$$
Differentiate the inner function:
$$
\frac{d}{dx}(3x^2)=6x,\quad \frac{d}{dx}\Bigl(2^{-6x}\Bigr)=2^{-6x}\ln(2)\cdot(-6)=-6\ln(2)\,2^{-6x}.
$$
Thus,
$$
f'(x)=40\Bigl(3x^2+2^{-6x}\Bigr)^{9}\Bigl(6x-6\ln(2)\,2^{-6x}\Bigr).
$$

b. We have 
$$
g(x)=15\ln\Bigl(-3e^{\frac{2}{3x}}\Bigr).
$$
Let 
$$
u(x)=-3e^{\frac{2}{3x}}.
$$
Then, by the chain rule,
$$
g'(x)=15\cdot\frac{u'(x)}{u(x)}.
$$
Differentiate $$u(x)$$:
$$
u'(x)=-3e^{\frac{2}{3x}}\cdot\frac{d}{dx}\Bigl(\frac{2}{3x}\Bigr).
$$
We have
$$
\frac{d}{dx}\Bigl(\frac{2}{3x}\Bigr)=-\frac{2}{3x^2},
$$
so
$$
u'(x)=-3e^{\frac{2}{3x}}\Bigl(-\frac{2}{3x^2}\Bigr)=\frac{2e^{\frac{2}{3x}}}{x^2}.
$$
Then,
$$
g'(x)=15\cdot\frac{\frac{2e^{\frac{2}{3x}}}{x^2}}{-3e^{\frac{2}{3x}}}=-15\cdot\frac{2}{3x^2}=-\frac{10}{x^2}.
$$

c. We have
$$
h(t)=\frac{1}{\sqrt[3]{(3t-4)^5}}+\ln(t^4).
$$
Rewrite the first term as
$$
(3t-4)^{-5/3},
$$
and note that 
$$
\ln(t^4)=4\ln(t).
$$
Differentiate term‑by‑term.

For the first term, using the chain rule,
$$
\frac{d}{dt}\Bigl[(3t-4)^{-5/3}\Bigr]=-\frac{5}{3}(3t-4)^{-8/3}\cdot(3)= -5(3t-4)^{-8/3}.
$$
For the second term,
$$
\frac{d}{dt}[4\ln(t)]=\frac{4}{t}.
$$
Thus,
$$
h'(t)=-5(3t-4)^{-8/3}+\frac{4}{t}.
$$

d. We have
$$
y=\frac{3}{5\Bigl(e^{12x+5}-6\Bigr)^3}+\pi^2.
$$
Since $$\pi^2$$ is constant, its derivative is zero. Rewrite the first term as
$$
\frac{3}{5}\Bigl(e^{12x+5}-6\Bigr)^{-3}.
$$
Differentiate using the chain rule:
$$
y'=\frac{3}{5}\cdot(-3)\Bigl(e^{12x+5}-6\Bigr)^{-4}\cdot\frac{d}{dx}\Bigl(e^{12x+5}-6\Bigr).
$$
Since
$$
\frac{d}{dx}(e^{12x+5})=12e^{12x+5},
$$
we have
$$
y'=-\frac{9}{5}\Bigl(e^{12x+5}-6\Bigr)^{-4}\cdot 12e^{12x+5}=-\frac{108\,e^{12x+5}}{5}\Bigl(e^{12x+5}-6\Bigr)^{-4}.
$$

e. We have
$$
f(m)=\frac{1}{\sqrt{12\ln(3m)}}.
$$
Rewrite as
$$
f(m)=(12\ln(3m))^{-1/2}.
$$
Differentiate using the chain rule:
$$
f'(m)=-\frac{1}{2}(12\ln(3m))^{-3/2}\cdot\frac{d}{dm}\Bigl(12\ln(3m)\Bigr).
$$
Differentiate the inner function:
$$
\frac{d}{dm}(12\ln(3m))=12\cdot\frac{1}{3m}\cdot 3= \frac{12}{m}.
$$
Thus,
$$
f'(m)=-\frac{1}{2}(12\ln(3m))^{-3/2}\cdot\frac{12}{m}=-\frac{6}{m}\,(12\ln(3m))^{-3/2}.
$$

f. We have
$$
g(x)=\Bigl(7^{\ln x}\Bigr)^2.
$$
Rewrite using properties of exponents:
$$
g(x)=7^{2\ln x}=e^{2\ln x\ln 7}=x^{2\ln 7}.
$$
Differentiate using the power rule:
$$
g'(x)=2\ln 7\cdot x^{2\ln 7-1}.
$$
a. $$ f'(x) = 40\left(3x^2 + 2^{-6x}\right)^9 \left(6x - 6\ln(2) \cdot 2^{-6x}\right) $$ b. $$ g'(x) = -\frac{10}{x^2} $$ c. $$ h'(t) = -5(3t - 4)^{-8/3} + \frac{4}{t} $$ d. $$ y' = -\frac{108e^{12x+5}}{5( e^{12x+5} - 6 )^4} $$ e. $$ f'(m) = -\frac{6}{m} \cdot (12\ln(3m))^{-3/2} $$ f. $$ g'(x) = 2\ln 7 \cdot x^{2\ln 7 - 1} $$

MATH 1020
Chain Rule
Learning Activity 4.3C
4. ( 6 pts ) Find the derivative of each of the following functions.
You need not take the time to simplify your derivatives.
a.

b.
c.
d.
e.
f.

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MATH 1020 Chain Rule Learning Activity 4.3C 4. ( 6 pts ) Find the derivative of each of the following functions. You need not take the time to simplify your derivatives. a. b. c. d. e. f.