Question

ii). $y = e^{-4x}$

iii). $f(x)=ln(7x)$

c. use product rule to differentiate $f(x)=x^{2}e^{x^{2}}$. you do not need to simplify.

d. use quotient rule to differentiate $g(x)=\frac{e^{2x}}{(x^{2}+1)^{10}}$

e. evaluate limit algebraically.

f. $lim_{x
ightarrow2}\frac{x^{2}-3x + 2}{x - 2}$

Explanation:

Step1: Differentiate $y = e^{-4x}$ using chain - rule

Let $u=-4x$, then $y = e^{u}$. $\frac{du}{dx}=-4$ and $\frac{dy}{du}=e^{u}$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. So $\frac{dy}{dx}=e^{-4x}\cdot(-4)=-4e^{-4x}$.

Step2: Differentiate $f(x)=\ln(7x)$ using chain - rule

Let $u = 7x$, then $f(x)=\ln(u)$. $\frac{du}{dx}=7$ and $\frac{df}{du}=\frac{1}{u}$. By the chain - rule $\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}=\frac{1}{7x}\cdot7=\frac{1}{x}$.

Step3: Differentiate $f(x)=x^{2}e^{x^{2}}$ using product rule

The product rule states that if $y = uv$, where $u$ and $v$ are functions of $x$, then $y^\prime=u^\prime v + uv^\prime$. Let $u = x^{2}$, $u^\prime=2x$; $v = e^{x^{2}}$, and by the chain - rule $v^\prime=e^{x^{2}}\cdot2x$. So $f^\prime(x)=2x\cdot e^{x^{2}}+x^{2}\cdot2x e^{x^{2}}=2xe^{x^{2}}(1 + x^{2})$.

Step4: Differentiate $g(x)=\frac{e^{2x}}{(x^{2}+1)^{10}}$ using quotient rule

The quotient rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Let $u = e^{2x}$, $u^\prime=2e^{2x}$; $v=(x^{2}+1)^{10}$, and by the chain - rule $v^\prime = 10(x^{2}+1)^{9}\cdot2x$. Then $g^\prime(x)=\frac{2e^{2x}(x^{2}+1)^{10}-e^{2x}\cdot20x(x^{2}+1)^{9}}{(x^{2}+1)^{20}}$.

Step5: Evaluate $\lim_{x

ightarrow2}\frac{x^{2}-3x + 2}{x - 2}$
Factor the numerator: $x^{2}-3x + 2=(x - 1)(x - 2)$. Then $\lim_{x
ightarrow2}\frac{x^{2}-3x + 2}{x - 2}=\lim_{x
ightarrow2}\frac{(x - 1)(x - 2)}{x - 2}=\lim_{x
ightarrow2}(x - 1)=1$.

Answer:

• For $y = e^{-4x}$, $\frac{dy}{dx}=-4e^{-4x}$
• For $f(x)=\ln(7x)$, $\frac{df}{dx}=\frac{1}{x}$
• For $f(x)=x^{2}e^{x^{2}}$, $f^\prime(x)=2xe^{x^{2}}(1 + x^{2})$
• For $g(x)=\frac{e^{2x}}{(x^{2}+1)^{10}}$, $g^\prime(x)=\frac{2e^{2x}(x^{2}+1)^{10}-e^{2x}\cdot20x(x^{2}+1)^{9}}{(x^{2}+1)^{20}}$
• For $\lim_{x

ightarrow2}\frac{x^{2}-3x + 2}{x - 2}$, the limit is $1$