practice exercises 19 - 60. derivatives find and si functions. 19. $f(x)=3x^{4}(2x^{2}-1)$ 21…

Explanation:

Step1: Solve for $f(x)=3x^{4}(2x^{2}-1)$

First, use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = 3x^{4}$ and $v=2x^{2}-1$. $u^\prime=12x^{3}$, $v^\prime = 4x$. $f^\prime(x)=12x^{3}(2x^{2}-1)+3x^{4}(4x)=24x^{5}-12x^{3}+12x^{5}=36x^{5}-12x^{3}$.

Step2: Solve for $f(x)=\frac{x}{x + 1}$

Use the quotient - rule $(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = x$, $u^\prime=1$, $v=x + 1$, $v^\prime = 1$. $f^\prime(x)=\frac{1\cdot(x + 1)-x\cdot1}{(x + 1)^{2}}=\frac{x + 1-x}{(x + 1)^{2}}=\frac{1}{(x + 1)^{2}}$.

Step3: Solve for $f(t)=t^{5/3}e^{t}$

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u=t^{5/3}$, $u^\prime=\frac{5}{3}t^{2/3}$, $v = e^{t}$, $v^\prime=e^{t}$. $f^\prime(t)=\frac{5}{3}t^{2/3}e^{t}+t^{5/3}e^{t}=t^{2/3}e^{t}(\frac{5}{3}+t)$.

Step4: Solve for $f(x)=\frac{e^{x}}{e^{x}+1}$

Use the quotient - rule $(\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = e^{x}$, $u^\prime=e^{x}$, $v=e^{x}+1$, $v^\prime=e^{x}$. $f^\prime(x)=\frac{e^{x}(e^{x}+1)-e^{x}\cdot e^{x}}{(e^{x}+1)^{2}}=\frac{e^{2x}+e^{x}-e^{2x}}{(e^{x}+1)^{2}}=\frac{e^{x}}{(e^{x}+1)^{2}}$.

Step5: Solve for $f(x)=xe^{-x}$

Use the product - rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x$, $u^\prime=1$, $v=e^{-x}$, $v^\prime=-e^{-x}$. $f^\prime(x)=1\cdot e^{-x}+x\cdot(-e^{-x})=e^{-x}(1 - x)$.

Answer:

For $f(x)=3x^{4}(2x^{2}-1)$, $f^\prime(x)=36x^{5}-12x^{3}$; For $f(x)=\frac{x}{x + 1}$, $f^\prime(x)=\frac{1}{(x + 1)^{[SSE Completed, Client Connection Error][LLM SSE On Failure]