Question

derivative xsinx principle method​

Answer 1

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

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Answer 2

Answer:

$sinx+xcosx$

Step-by-step explanation:

$f(x)=xsinx\\\\WKT\\f^I(x)=Lt_{h\to0}\frac{f(x+h)-f(x)}{h}\\\\f^I(x)=Lt_{h\to0}\frac{(x+h)sin(x+h)-xsinx}{h}\\\\f^I(x)=Lt_{h\to0}(sin(x+h)+\frac{xsin(x+h)-xsinx}{h})\\\\f^I(x)=sinx+Lt_{h\to0}(\frac{xsinxcosh+xsinhcosx-xsinx}{h})\\\\Since\;\;sin(a+b)=sinacosb+cosasinb\;\;and \;\;Lt_{h\to0}\frac{sinh}{h}=1\;\;and \frac{cosh-1}{h}=0\\\\f^I(x)=sinx+Lt_{h\to0}(\frac{xsinhcosx}{h}+\frac{xsinx(cosh-1)}{h})\\\\f^I(x)=sinx+xcosx$

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