Using an appropriate form of the chain rule, find all (partial) derivaitves of the 1st order of ...

Question

... $f(g(x,y))$, where $f(z)=\ln(1+z)$ and $g(x,y)=\sqrt{x^2+y^2}$.

Please help, very confused about this question. Thanks.

Answer 1

If $f(z)=\ln(1+z)$ and $g(x,y)=\sqrt{x^2+y^2}$, then

$f(g(x,y))=\ln(1+\sqrt{x^2+y^2})$

Then, using the fact that $\frac{d}{dx}(\ln f(x))=\frac{f'(x)}{f(x)}$, where $f'(x)=\frac{df(x)}{dx}$, we have

$\large \frac{\partial f(g(x,y))}{\partial x}=\frac{\frac{\partial}{\partial x}(1+\sqrt{x^2+y^2})}{1+\sqrt{x^2+y^2}}$

Invoking the chain rule, we have

$\large\frac{\partial}{\partial x}(1+\sqrt{x^2+y^2})=\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{\sqrt{x^2+y^2}}$

Thus we have

$\large \frac{\partial f(g(x,y))}{\partial x}=\frac{x}{\sqrt{x^2+y^2}(1+\sqrt{x^2+y^2})}$

In order to evaluate the partial derivative with respect to $y$, going through a similar procedure, we have

$\large \frac{\partial f(g(x,y))}{\partial y}=\frac{\frac{\partial}{\partial y}(1+\sqrt{x^2+y^2})}{1+\sqrt{x^2+y^2}}=\frac{y}{\sqrt{x^2+y^2}(1+\sqrt{x^2+y^2})}$

Answer 2

We know that function $f(g(x,y))$. Thus, $$\dfrac{\partial f(g(x,y))}{\partial x}=\dfrac{\partial \ln(1+\sqrt{x^2+y^2})}{\partial x}=\\ \dfrac{\partial \ln(1+\sqrt{x^2+y^2})}{\partial (1+\sqrt{x^2+y^2})}\dfrac{\partial (1+\sqrt{x^2+y^2})}{\partial x}=\dfrac{x}{(1+\sqrt{x^2+y^2})\sqrt{x^2+y^2}}$$ Also, $$\dfrac{\partial f(g(x,y))}{\partial y}=\dfrac{\partial \ln(1+\sqrt{x^2+y^2})}{\partial (1+\sqrt{x^2+y^2})}\dfrac{\partial (1+\sqrt{x^2+y^2})}{\partial y}=\dfrac{y}{(1+\sqrt{x^2+y^2})\sqrt{x^2+y^2}}$$