1. find the derivative of each function\n2. find the following derivatives\na) $\frac{d}{dx}(x^{3}\tan^{-1}(5…

Explanation:

Step1: Recall product - rule and chain - rule

The product - rule states that if $y = u\cdot v$, then $y^\prime=u^\prime v + uv^\prime$. The chain - rule states that if $y = f(g(x))$, then $y^\prime=f^\prime(g(x))\cdot g^\prime(x)$. Also, $\frac{d}{dx}\tan^{- 1}(u)=\frac{1}{1 + u^{2}}\cdot\frac{du}{dx}$, $\frac{d}{dx}e^{u}=e^{u}\cdot\frac{du}{dx}$, $\frac{d}{dx}\arcsin(u)=\frac{1}{\sqrt{1 - u^{2}}}\cdot\frac{du}{dx}$.

Step2: Solve part a

Let $u = x^{3}$ and $v=\tan^{-1}(5x)$. Then $u^\prime = 3x^{2}$ and $v^\prime=\frac{1}{1+(5x)^{2}}\cdot5=\frac{5}{1 + 25x^{2}}$. By the product - rule, $\frac{d}{dx}(x^{3}\tan^{-1}(5x))=3x^{2}\tan^{-1}(5x)+x^{3}\cdot\frac{5}{1 + 25x^{2}}=3x^{2}\tan^{-1}(5x)+\frac{5x^{3}}{1 + 25x^{2}}$.

Step3: Solve part b

Let $u=\sqrt{x + 1}=(x + 1)^{\frac{1}{2}}$. Then $u^\prime=\frac{1}{2}(x + 1)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x+1}}$. By the chain - rule, $\frac{d}{dx}e^{\sqrt{x + 1}}=e^{\sqrt{x + 1}}\cdot\frac{1}{2\sqrt{x + 1}}=\frac{e^{\sqrt{x+1}}}{2\sqrt{x + 1}}$.

Step4: Solve part c

Let $u = 4x^{2}-1$. Then $u^\prime=8x$. By the chain - rule, $\frac{d}{dx}\arcsin(4x^{2}-1)=\frac{8x}{\sqrt{1-(4x^{2}-1)^{2}}}=\frac{8x}{\sqrt{1-(16x^{4}-8x^{2}+1)}}=\frac{8x}{\sqrt{8x^{2}-16x^{4}}}=\frac{8x}{2x\sqrt{2 - 4x^{2}}}=\frac{4}{\sqrt{2 - 4x^{2}}}$ for $x\neq0$. When $x = 0$, the derivative is $0$.

Answer:

a) $3x^{2}\tan^{-1}(5x)+\frac{5x^{3}}{1 + 25x^{2}}$ b) $\frac{e^{\sqrt{x+1}}}{2\sqrt{x + 1}}$ c) $\frac{4}{\sqrt{2 - 4x^{2}}}$ (for $x\neq0$), $0$ (for $x = 0$)