Question

differentiate xsinx by definition​

Answer 1

Answer:

$sinx+xcosx$

Step-by-step explanation:

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

Hope it Helps

Answer 2

Answer:

sinx+xcosxsinx+xcosx

Step-by-step explanation:

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