Question

find the derivative of the function. y = sin^(-1)(8x + 1) y = blank resources read it watch it master it submit answer 18. - / 1 points find the derivative of the function. y = 5 tan^(-1)(x - sqrt(1 + x^2)) y = blank resources read it

Explanation:

Step1: Recall derivative of inverse - sine function

The derivative of $y = \sin^{-1}(u)$ with respect to $x$ is $\frac{dy}{dx}=\frac{1}{\sqrt{1 - u^{2}}}\cdot\frac{du}{dx}$ by the chain - rule. Here $u = 8x+1$, and $\frac{du}{dx}=8$.

Step2: Apply the chain - rule

Substitute $u = 8x + 1$ and $\frac{du}{dx}=8$ into the formula. We get $y'=\frac{8}{\sqrt{1-(8x + 1)^{2}}}$.

for second part:

Step1: Recall derivative of inverse - tangent function

The derivative of $y = \tan^{-1}(u)$ with respect to $x$ is $\frac{dy}{dx}=\frac{1}{1 + u^{2}}\cdot\frac{du}{dx}$ by the chain - rule. Here $u=x-\sqrt{1 + x^{2}}$. First, find $\frac{du}{dx}$.
The derivative of $x$ with respect to $x$ is $1$, and for $v=\sqrt{1 + x^{2}}=(1 + x^{2})^{\frac{1}{2}}$, using the chain - rule $\frac{dv}{dx}=\frac{1}{2}(1 + x^{2})^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{1 + x^{2}}}$. So $\frac{du}{dx}=1-\frac{x}{\sqrt{1 + x^{2}}}=\frac{\sqrt{1 + x^{2}}-x}{\sqrt{1 + x^{2}}}$.

Step2: Apply the chain - rule

Since $y = 5\tan^{-1}(u)$, then $y'=5\cdot\frac{1}{1+(x-\sqrt{1 + x^{2}})^{2}}\cdot\frac{\sqrt{1 + x^{2}}-x}{\sqrt{1 + x^{2}}}$.
Simplify the denominator $1+(x-\sqrt{1 + x^{2}})^{2}=1+x^{2}-2x\sqrt{1 + x^{2}}+1 + x^{2}=2 + 2x^{2}-2x\sqrt{1 + x^{2}}$.
Also, note that $y'=\frac{5}{\sqrt{1 + x^{2}}}\cdot\frac{\sqrt{1 + x^{2}}-x}{2 + 2x^{2}-2x\sqrt{1 + x^{2}}}=\frac{5}{2\sqrt{1 + x^{2}}}$.

Answer:

$\frac{8}{\sqrt{1-(8x + 1)^{2}}}$