I need help to find derivative of: $\frac{d}{dx}(x+1)^{\sin x}$
i tried to do something like this..
$$(x+1)^{\sin x}\cdot \ln\left(x+1\right)=\sin x(x+1)^{\sin\left(x\right)-1}\cdot \ln(x+1)-(x+1)^{\sin(x)}\cdot \frac{1}{x+1}\:=\sin x(x+1)^{\sin\left(x\right)-1} \ln(x+1)-\frac{(x+1)^{\sin x}}{x+1}$$
I tried another way:
$\left(x+1\right)^{sinx}\:=\:ln\left(x+1\right)^{sinx}\:=\:sin\left(x\right)ln\left(x+1\right)\:=\:cos\:\cdot \:ln\left(x+1\right)+sin\left(x\right)\cdot \frac{1}{x+1}\:=\:cos\left(x\right)ln\left(x+1\right)\:+\:\frac{sin\left(x\right)}{x+1}$
ok i got the solution! I put it there, maybe this will help someone!
i used this rule : $e^a=e^{a\cdot ln\cdot e}$
$\left(x+1\right)^{sinx}=\:e^{sinx\cdot ln\left(x+1\right)}=\:e^{sinx\cdot ln\left(x+1\right)}\cdot cosx\:\cdot \:ln\left(x+1\right)\:+\:sinx\cdot \frac{1}{x+1}\:=\:\left(x+1\right)\left(cosxln\left(x+1\right)\:+\:\frac{sinx}{x+1}\right)$
You have:
$$f(x) = (x+1)^{\sin{x}},$$ so $\ln{f} = \sin{x} \, \ln{(x+1)}$. Now we have:
$$ \frac{d}{dx} \ln {f} = \frac{1}{f} \frac{df}{dx} = \frac{d}{dx}\left[ \sin{x} \, \ln{(x+1)} \right] . $$
You can now solve for $f'$ once you expand the LHS of the equation.
Cheers!