Question

Find the derivative of xsinx from 1st principle

Answer 1

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Answer 2

Answer:

$\frac{d}{dx}(x\sin x)=x\cos x+\sin x$

Step-by-step explanation:

Given : Expression $x\sin x$

To find : The derivative of given expression from first principle?

Solution :  

The derivative rule of first principal is  

$\frac{d}{dx}(a\cdot b)=a\times \frac{d}{dx}(b)+b\times \frac{d}{dx}(a)$

Here, a=x and b=sin x

$\frac{d}{dx}(x\sin x)=x\times \frac{d}{dx}(\sin x)+\sin x\times \frac{d}{dx}(x)$

$\frac{d}{dx}(x\sin x)=x\times (\cos x)+\sin x\times 1$

$\frac{d}{dx}(x\sin x)=x\cos x+\sin x$

Therefore,  $\frac{d}{dx}(x\sin x)=x\cos x+\sin x$