Question

if y= sin inverse [ 5x +12 root 1-x2 by 13 find dy by dx

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given:

$y=sin^{-1}(\frac{5x+12\sqrt{1-x^2}}{13})$

To find :

$\frac{dy}{dx}$

Solution:

$y=sin^{-1}(\frac{5x+12\sqrt{1-x^2}}{13})$

$y=sin^{-1}(\frac{5}{13}x+\frac{12}{13}\sqrt{1-x^2})$

$y=sin^{-1}(x\sqrt{\frac{25}{169}}+\frac{12}{13}\sqrt{1-x^2})$

$y=sin^{-1}(x\sqrt{1-(\frac{12}{13})^2}+\frac{12}{13}\sqrt{1-x^2})$

We know that

$sin^{-1}x+sin^{-1}y=sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2})$

$y=sin^{-1}x+sin^{-1}(\frac{12}{13})$

Now, differentiate w.r.t x

$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+0$

By using the formula

$\frac{d(sin^{-1}(x))}{dx}=\frac{1}{\sqrt{1-x^2}}$

$\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$