MHB Partial Derivatives: Solving Difficult Problems

[Kamo123]

Hello

I'm currently trying to solve these two problems:

1) Find the partial derivatives ∂m/∂q and ∂m/∂h of the function:

m=ln(qh-2h^2)+2e^(q-h^2+3)^4-7

Here, I know I should differentiate m with respect to q while treating h as a constant and vice versa. But I'm still stuck, and I'm not sure how to actually do it.

2) Find the partial derivative ∂z/∂x of the function:

(z+1)^3y-(zx)^2-3=x/z-ln(xyz)-5xy^2

I have tried some implicit differentiation here, but I can't really make it work out.

Please explain how to solve the two problems very explicitly.

 

[Ackbach]

Hello

I'm currently trying to solve these two problems:

1) Find the partial derivatives ∂m/∂q and ∂m/∂h of the function:

m=ln(qh-2h^2)+2e^(q-h^2+3)^4-7

Here, I know I should differentiate m with respect to q while treating h as a constant and vice versa. But I'm still stuck, and I'm not sure how to actually do it.

So you have
$$m=\ln(qh-2h^2)+2e^{(q-h^2+3)^4}-7.$$
Is this correct?

2) Find the partial derivative ∂z/∂x of the function:

(z+1)^3y-(zx)^2-3=x/z-ln(xyz)-5xy^2

I have tried some implicit differentiation here, but I can't really make it work out.

Please explain how to solve the two problems very explicitly.

So here you have
$$(z+1)^{3y}-(zx)^2-3=\frac{x}{z}-\ln(xyz)-5xy^2.$$
Is that correct?

 

[Kamo123]

So you have
$$m=\ln(qh-2h^2)+2e^{(q-h^2+3)^4}-7.$$
Is this correct?

So here you have
$$(z+1)^{3y}-(zx)^2-3=\frac{x}{z}-\ln(xyz)-5xy^2.$$
Is that correct?

1) Yes :-)

2) Not exactly. Here you have it:

 

[Ackbach]

Ok, so for the first one, you have
\begin{align*}
\pd{m}{q}&=\pd{}{q}\left[\ln(qh-2h^2)+2e^{(q-h^2+3)^4}-7\right] \\
&=\frac{1}{qh-2h^2} \, \pd{}{q}(qh-2h^2)+2e^{(q-h^2+3)^4}\,\pd{}{q}\left[(q-h^2+3)^4\right].
\end{align*}
Can you continue? All the normal rules apply: products, quotients, and composition.

For the second one, we use the same procedure we use for functions of a single variable. Here we are assuming that $z=z(x,y)$, and hence we compute:
$$\pd{}{x}\left[ (z+1)^3 y-(zx)^2-3=\frac{x}{z}-\ln(xyz)-5xy^2 \right],$$
or
$$3(z+1)^2y\pd{z}{x}-(2zx^2\pd{z}{x}+2z^2x)=\frac{z-x\pd{z}{x}}{z^2}-\frac{1}{xyz}\pd{(xyz)}{x}-5y^2.$$
Can you continue?