Question

How do you find the derivative of `3x^(lnx)`?

Answer 1

Let `y=3x^{ln(x)}`, then `ln(y)=ln(3x^{ln(x)})=ln(3)+ln(x^{ln(x)})=ln(3)+(ln(x))^{2}` (using the properties that `ln(AB)=ln(A)+ln(B)` for `A,B > 0` and `ln(A^{p})=pln(A)` for `A > 0`).

Now differentiate `\ln(y)=ln(3)+(ln(x))^{2}` with respect to `x`, keeping in mind that `y` is a function of `x` and using the Chain Rule to get `\frac{1}{y}\cdot \frac{dy}{dx}=2(ln(x))^{1}\cdot \frac{1}{x}=\frac{2\ln(x)}{x}`.

Multiplying both sides of this last equation by `y=3x^{ln(x)}` helps us see that `\frac{dy}{dx}=3x^{ln(x)}\cdot \frac{2\ln(x)}{x}=6x^{ln(x)-1}ln(x)`. This is valid as long as `x > 0`.