Question

let (y = \tan^{-1}(sqrt{5x^{2}-1})), then (\frac{dy}{dx}=)

Explanation:

Step1: Recall the chain - rule

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 $u=\sqrt{5x^{2}-1}$.

Step2: Find $\frac{du}{dx}$

If $u=(5x^{2}-1)^{\frac{1}{2}}$, then by the power - rule and chain - rule, $\frac{du}{dx}=\frac{1}{2}(5x^{2}-1)^{-\frac{1}{2}}\cdot(10x)=\frac{5x}{\sqrt{5x^{2}-1}}$.

Step3: Substitute $u$ and $\frac{du}{dx}$ into the chain - rule formula

Since $y = \tan^{- 1}(u)$ and $u=\sqrt{5x^{2}-1}$, we have $\frac{dy}{dx}=\frac{1}{1+(5x^{2}-1)}\cdot\frac{5x}{\sqrt{5x^{2}-1}}$.

Step4: Simplify the expression

First, simplify the denominator of the first fraction: $1+(5x^{2}-1)=5x^{2}$. Then $\frac{dy}{dx}=\frac{1}{5x^{2}}\cdot\frac{5x}{\sqrt{5x^{2}-1}}=\frac{1}{x\sqrt{5x^{2}-1}}$ for $x
eq0$.

Answer:

$\frac{1}{x\sqrt{5x^{2}-1}}$