What is the derivative of $\log_x(A)$ where $x$ is the base (differentiaition with respect to $x$) I want to find out what $\frac{d}{dx}\log_x A$ is?  I did this so far but I'm not sure.  $y = \log_x A \Longrightarrow x^y = A$ so, $d/dx(x^y) = d/dx(A)$ [differentiating both sides w.r.t $x$] then, $y\cdot x^{y-1}dy/dx = 0$ ....that would imply that $dy/dx = 0$ ..what am i doing wrong? as the graph of $\log_x A$ is a curve. 
 A: Use the identity $$\log_x(A)=\frac{\ln(A)}{\ln(x)}.$$ Thus, $$\frac{d}{dx}\log_x(A)=\frac{d}{dx}\frac{\ln(A)}{\ln(x)}=-\frac{\ln(A)}{x\ln^2(x)}.$$
A: If $y = \log_x A$ then $y=(\log_A x)^{-1}$ so $$\dfrac{dy}{dx} = -(\log_A x)^{-2}\cdot \dfrac d {dx} \log_A x = -(\log_A x)^{-2} \cdot \dfrac 1 x \cdot \dfrac 1 {\log_e A}.$$
However, suppose one writes this in the form
$$
x^y = \text{constant}.
$$
Then to find the derivative of this with respect to $x$, one must realize that $x$ and $y$ are both changing as functions of $x$.  One has
$$
\frac d {dx} f(x)^{g(x)} = g(x) f(x)^{g(x)-1} g'(x) + f(x)^{g(x)}(\log_e f(x)) f'(x).
$$
One way to show that is by logarithmic differentiation.  So
$$
\frac d{dx} x^y = yx^{y-1}\frac {dy}{dx} + x^y(\log_e x)\frac{dx}{dx} = yx^{y-1} \frac{dy}{dx} + x^y \log_e x.
$$
Or you could approach the whole thing by logarithmic differentiation in the first place:
\begin{align}
\log_e (x^y) & = \text{constant (which is the logarithm of the earlier constant)} \\[10pt]
y \log_e x & = \text{constant} \\[10pt]
y \cdot \frac 1 x + \frac{dy}{dx}\cdot \log_e x & = 0 \\[10pt]
\frac{dy}{dx} & = \frac{-y}{x\log_e x} = \frac{-\log_x A}{x\log_e x}.
\end{align}
