How would I do this partial derivative?
$\displaystyle \frac {d^2z}{dydx} ...........z= \frac{(x+y)^{y+1}}{y+1} +sinh^3(y)+2 $
Thanks.
Hello,
Firstly, derivate in function of x.
The derivative of $\displaystyle \frac{(x+y)^{y+1}}{y+1}$ is $\displaystyle (x+y)^y$ (do you know why ?)
The derivative of $\displaystyle sinh^3(y)+2$ is 0 because there is no x in it, so it's like a constant if the variable is x.
Now remains $\displaystyle (x+y)^y$ to derivate in function of y.
$\displaystyle (x+y)^y=e^{y \ln(x+y)}$
The derivative for $\displaystyle e^{f(x)}$ is $\displaystyle f'(x)e^{f(x)}$
Here, $\displaystyle f(y)=y \ln(x+y)$. So, by using the product rule, we have : $\displaystyle f'(y)=\frac{y}{x+y}+\ln(x+y)$
Hence $\displaystyle \frac{d^2z}{dydx}=e^{y \ln(x+y)}(\frac{y}{x+y}+\ln(x+y))$$\displaystyle =(x+y)^y \frac{y}{x+y}+(x+y)^y \ln(x+y)=y(x+y)^{y-1}+(x+y)^y \ln(x+y)$