Find the partial derivative

Question

How to find the partial derivative of this function, with respect for x? $$ h(x,y)=\frac{1}{\ln(e^x+y)}, y>0 $$

Answer 1

The way I teach students to deal with partial differentiation, in order to keep the "accounting" straight, is to replace with placeholder constant(s) everything involving the variable(s) you are not differentiating with respect to, and handle the resulting expression as if it were plain old single-variable differentiation. (All of the rules you've already learned there still apply.) Thus, treat

$$ \frac{\partial h}{\partial x} \ \ \text{as} \ \ \frac{d}{dx} \ \left[ \frac{1}{\ln(e^x + C)} \right] \ , $$

and then restore the " $ y $ " appropriately in the expression for the derivative. Similarly, write

$$ \frac{\partial h}{\partial y} \ \ \text{as} \ \ \frac{d}{dy} \ \left[ \frac{1}{\ln(D + y)} \right] \ , $$

and then restore the " $ e^x $ " after completing the "$ y-$ differentiation".

Answer 2

Set $u = e^x + y$ and use the chain rule, taking note that

$h(x, y) = (\ln u)^{-1,} \tag{1}$

whence

$\dfrac{\partial h(x, y)}{\partial x} = \dfrac{dh}{du}\dfrac{\partial u}{\partial x}; \tag{2}$

we have

$\dfrac{dh}{du} = -(\ln u)^{-2} u^{-1} \tag{3}$

and

$\dfrac{\partial u(x, y)}{\partial x} = e^x, \tag{4}$

so bringing it all together yields

$\dfrac{\partial h(x, y)}{\partial x} = -(\ln(e^x + y))^{-2} (e^x + y)^{-1}e^x; \tag{5}$

likewise, since

$\dfrac{\partial u(x, y)}{\partial y} = 1, \tag{6}$

$\dfrac{\partial h(x, y)}{\partial y} = -(\ln(e^x + y))^{-2} (e^x + y)^{-1}. \tag{7}$

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!