$$\text { Let } y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1 < x<1 \text {. Then at } x=\frac{1}{2} \text {, the value of } 225\left(y^{\prime}-y^{\prime \prime}\right) \text { is equal to }$$
Solution
<p>$$\begin{aligned}
& y=\log _e\left(\frac{1-x^2}{1+x^2}\right) \\
& \frac{d y}{d x}=y^{\prime}=\frac{-4 x}{1-x^4}
\end{aligned}$$</p>
<p>Again,</p>
<p>$$\frac{d^2 y}{d x^2}=y^{\prime \prime}=\frac{-4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$$</p>
<p>Again</p>
<p>$$y^{\prime}-y^{\prime \prime}=\frac{-4 x}{1-x^4}+\frac{4\left(1+3 x^4\right)}{\left(1-x^4\right)^2}$$</p>
<p>at $\mathrm{x}=\frac{1}{2}$,</p>
<p>$y^{\prime}-y^{\prime \prime}=\frac{736}{225}$</p>
<p>Thus $225\left(y^{\prime}-y^{\prime \prime}\right)=225 \times \frac{736}{225}=736$</p>
About this question
Subject: Mathematics · Chapter: Differentiation · Topic: Derivatives of Standard Functions
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