# Implicit Differentiation of logarithm

Differentiate $$y=\log_a(x)$$ with respect to $$x$$

I see that $$a^y=x$$. My textbook says implicit differentiation gets us \begin{align*}a'(\ln a)\frac{dy}{dx}&=1 \\\implies \frac{dy}{dx}&=\frac{1}{a'\ln a} \\ \frac{dy}{dx}&=\frac{1}{x\ln a}\end{align*}

What I don't understand is why $$\frac{d}{dx}[a^y]=a'(\ln a)\cfrac{dy}{dx}$$ and why $$a'=x$$ When I try this using a base of $$e$$ with the chain rule, I get \begin{align*}\frac{d}{dx}[e^{y\ln a}]&=\frac{d}{dx}[x] \\ &\boxed{u=y\ln a, du=\frac{dy}{dx}\ln a+\frac1ay; \\ f=e^u, df=e^u \\ df/du*du/dx=e^{y\ln a}\frac{dy}{dx}\ln a+\frac1ay} \\ \implies x\frac{dy}{dx}\ln a+\frac1ay&=1 \\ \frac{dy}{dx}&=\frac{1}{\ln a}\biggr(\frac1x-\frac{y}{a}\biggr)\end{align*}

I see here that if I distribute, I get $$\cfrac{1}{x\ln a}-\cfrac{y}{a\ln a}$$ which implies $$y$$ must be zero! But I don't know how to show that, either. Can someone fill the gaps I'm missing in my textbooks solution?

UPDATE: I just realized the mistake I made in my differentiation was forgetting that ln (a) is a constant! Once I took out the constant or allowed the constant to be differentiated to $$0$$ I got the correct answer. I will mark the best answer correct soon enough, though, thanks everyone

• $(a^y)'=(e^{y\ln a})'=(e^{y\ln a})y' \ln a=a^yy' \ln a$ Commented Mar 2, 2020 at 20:11
• That first $a'$ should be $a^y$. Then $x$ is substituted back in for it. Commented Mar 2, 2020 at 20:15

An alternative derivation avoids implicit differentiation altogether. Note that $$y = \log_a x = \frac{\ln x}{\ln a} = \frac{1}{\ln a} \cdot \ln x$$ Since $$a$$ is constant, $$y' = \frac{1}{\ln a} \cdot \frac{1}{x} = \frac{1}{x \ln a}$$

$$a^y=x$$ differentiates on $$x$$ as

$$(a^y)'=1$$

and by the chain rule,

$$(a^y)'=\frac{d\,a^y}{dx}=\log(a)\,a^y\frac{dy}{dx}=\log(a)\,x\,y'.$$

Option:

$$a^y=x$$;

Tale $$\log_e$$ of both sides:

$$y \log a=\log x$$;

Differentiate with respect to $$x$$:

$$y' \log a=\dfrac{1}{x}$$;

$$y'=\dfrac{1}{x \log a }$$;

You can differentiate wrt $$y$$: $$\frac{d}{dx}[a^y]=\frac{d}{dy}[a^y] \frac {dy}{dx}=\frac{d} {dy}[a^y] y'=\frac{d}{dy}[e^{y \ln a}] y'$$ Note that $$(e^{cx})'=e^{cx}c$$ $$\frac{d}{dx}[a^y]=[e^{y \ln a}] \ln |a| y'$$ And $$e^{y \ln a}=a^y$$: $$\frac{d}{dx}[a^y]=a^y \ln |a| y'$$ Therefore: $$a^y=x \implies a^y \ln |a| y' =1$$ $$x \ln |a| y' =1 \implies y'= \frac {1}{x \ln |a|}$$