Calculate derivative of
$$g(x)=cos(x)exp(x^{2})$$

Question

Calculate derivative of 
$$g(x)=cos(x)exp(x^{2})$$

geerky42 · Answer

Apply product rule.

You take derivative of $\cos x$, leave $\exp(x^2)$ untouched, then you add it by derivative of $\exp(x^2)$ and leave $\cos x$ untouched.

While taking derivative of $\exp(x^2)$, apply chain rule there.

anonymous · Answer

chain rule was $$f(g(x)) \rightarrow f^{'}(g(x)).g^{'}(x)$$ right ?

turingtest · Answer

yes

geerky42 · Answer

So you figured it out?

anonymous · Answer

Is second step $$f^{'}(co(x)exp(x^{2}).(-sin(x).exp(2x))$$ ?

geerky42 · Answer

No, it's $-\sin(x)\cdot\exp(x^2) + \cos(x)\cdot2x\exp(x^2)$

anonymous · Answer

ok then i need to use this rule right  ? $$fg=fg^{'}+f^{'}g$$

geerky42 · Answer

Derivative of $\cos (x) $ is $-\sin(x)$,  and derivative of $\exp(x^2)$ is $2x\exp(x^2)$

So $f'(x)$ would be $(\cos(x))'\cdot\exp(x^2)+\cos(x)\cdot(\exp(x^2))'\~\=−\sin(x)⋅\exp(x^2)+\cos(x)⋅2x\exp(x^2)$

geerky42 · Answer

Yes, that rule.

geerky42 · Answer

Refresh page if you see question marks, lol

anonymous · Answer

ok i got it but only my problem is how to figure out when we which rule apply

turingtest · Answer

basically, when taking the derivative of the product of two functions, use the product rule$$D_x[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)$$when taking the derivative of a composite function use the chain rule$$D_x(f[g(x)])=f'[g(x)]g'(x)$$

anonymous · Answer

ok TuringTest now its better to understand thank you and geerky both

anonymous · Answer

Write the binary number $$11,101\overline{110}$$ as decimal fraction and explain it with help of geometric series.

turingtest · Answer

please post each question in a separate thread