If y=1+x^y+sin^(-1)sin^2x, then dy/dx at x=0 is

Correct answer: (C) 1. dy/dx=1+x^yy/(1+x)+ln⁡1+x⋅dy/dx+sin2x/(√(1-sin^4x)) Putting x=0 & y=1, we get, dy/dx=11+ln⁡1.dy/dx+0=1

Q18 / 46 · NTA Abhyas (JEE Main) 2020 · Mathematics · Differentiation

If $y = {(1 + x)}^{y} + s i n^{- 1} (s i n^{2} x)$ , then $\frac{d y}{d x}$ at $x = 0$ is

A
$0$
B
$\ln 2$
Correct answer
C
$1$
D
$\frac{1}{2}$

Solution

$\frac{d y}{d x} = {(1 + x)}^{y} \{\frac{y}{1 + x} + \ln (1 + x) \cdot \frac{d y}{d x}\} + \frac{s i n 2 x}{\sqrt{1 - s i n^{4} (x)}}$
Putting $x = 0$ & $y = 1$ , we get,

$\frac{d y}{d x} = 1 \{1 + \ln (1) . \frac{d y}{d x}\} + 0 = 1$

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

If y=1+xy+sin-1sin2x, then dydx at x=0 is

Solution

More Differentiation Questions

If $y = {(1 + x)}^{y} + s i n^{- 1} (s i n^{2} x)$ , then $\frac{d y}{d x}$ at $x = 0$ is