Answer to Question #108355 in Calculus for Nimra

Question

Differentiate F(x)= arc sinx . sinx²

Answer 1

Consider the function $F(x)=\arcsin(x)\cdotp \sin(x^2)$

Use product rule to differentiate the function as follows:

$$F'(x)=\frac{d}{dx}[\arcsin(x)\cdotp \sin(x^2)]$$

$$F'(x)=\arcsin(x)\frac{d}{dx}(\sin(x^2))+(\sin(x^2))\frac{d}{dx}(\arcsin(x))$$

Using chain rule, the derivative of $\sin(x^2)$ with respect to $x$ is evaluated as,

$$\frac{d}{dx}(\sin(x^2))=\cos(x^2)\cdotp \tfrac{d}{dx}(x^2)=2x\cos(x^2)$$

The derivative of $\arcsin(x)$ with respect to $x$ is evaluated as,

$$\tfrac{d}{dx}(\arcsin(x))=\tfrac{1}{\sqrt{1-x^2}}$$

Now, substitute the derivatives to obtain,

$$F'(x)=\arcsin(x)\cdotp2x\cos(x^2)+(\sin(x^2))\cdotp\tfrac{1}{\sqrt{1-x^2}}$$

Therefore, the derivative of the function is

$$F'(x)=2x\cos(x^2)\cdotp\arcsin(x)+\tfrac{\sin(x^2)}{\sqrt{1-x^2}}$$