What is the derivative of $\log_x(A)$ where $x$ is the base (differentiaition with respect to $x$)

Question

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.

Answer 1

Use the identity $$\log_x(A)=\frac{\ln(A)}{\ln(x)}.$$ Thus, $$\frac{d}{dx}\log_x(A)=\frac{d}{dx}\frac{\ln(A)}{\ln(x)}=-\frac{\ln(A)}{x\ln^2(x)}.$$

Answer 2

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}