Q106 / 160 · TS EAMCET 2018 (05 May Shift 1) · Mathematics

The slope of the tangent to the curve $f(x)=\tanh ^{-1}(\sin x)$ at $x=\pi$ is

  • A
    1
  • B
    0
  • Correct answer
    C
    -1
  • D
    -2

Solution

We have, $$ f(x)=\tanh ^{-1}(\sin x) $$ Let $$ \begin{aligned} f(x) & =y \\ y & =\tanh ^{-1}(\sin x) \\ \tanh y & =\sin x \end{aligned} $$ Differentiate w.r.t. ' $x$ ' $$ \begin{aligned} \operatorname{sech}^2 y \frac{d y}{d x} & =\cos x \quad \frac{d y}{d x}=\frac{\cos x}{\operatorname{sech}^2 y} \\ \frac{d y}{d x} & =\frac{\cos x}{1-\tanh ^2 y} \quad\left[\because \operatorname{sech}^2 y=1-\tanh ^2 y\right] \end{aligned} $$ At $$ \left.\frac{d y}{d x}\right|_{x=\pi} ^x=\frac{(-1)}{1-(0)} $$ [ $\because$ [from Eq. (i)] $\tanh y=\sin x, x=\pi \tanh y=0$ ] $$ \left.\frac{d y}{d x}\right|_{x=\pi}=-1 $$ Hence, slope of tangent to the curve is -1 .

Hence, the correct answer is option: 'C'.

Relevant Questions