Question

How do you find the derivative for `y=(sin(3x))^(ln(x))`?

Answer 1

Use logarithmic differentiation. Take the natural logarithm of both sides of `y=(sin(3x))^{ln(x)}` to get `ln(y)=ln((sin(3x))^{ln(x)})`. Then use the property `ln(a^{b})=bln(a)` (for `a>0`) to write `ln(y)=ln(x)ln(sin(3x))`.

Now differentiate both sides of this last equation with respect to `x`, keeping in mind that `y` is a function of `x` and using the Chain Rule and the Product Rule:

`1/y * dy/dx = 1/x * ln(sin(3x))+ln(x)* 1/(sin(3x)) * 3cos(3x).`

Simplifying a bit and multiplying both sides of this last equation by `y=(sin(3x))^{ln(x)}` helps us conclude that

`dy/dx =(sin(3x))^{ln(x)}*(\frac{ln(sin(3x))}{x}+3ln(x)*cot(3x))`