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

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To find the derivative of (y = (\sin(3x))^{\ln(x)}), we'll use the chain rule and the properties of logarithmic and exponential functions. Here's the process:

- Let (u = \sin(3x)) and (v = \ln(x)).
- Apply the chain rule to find (\frac{du}{dx}) and (\frac{dv}{dx}).
- Use the formula for differentiating (f(g(x))) where (f) is a function of (g(x)).
- Substitute (u = \sin(3x)), (v = \ln(x)), (\frac{du}{dx}), and (\frac{dv}{dx}) into the formula.
- Compute the derivative ( \frac{dy}{dx} ) by multiplying ( \frac{dy}{du} ) and ( \frac{du}{dx} ), and then adding the product of ( \frac{dy}{dv} ) and ( \frac{dv}{dx} ).

Let's go through the steps to find ( \frac{dy}{dx} ):

- (u = \sin(3x)) implies ( \frac{du}{dx} = 3\cos(3x) ).
- (v = \ln(x)) implies ( \frac{dv}{dx} = \frac{1}{x} ).
- (y = u^v) implies ( \frac{dy}{du} = vu^{v-1} ) and ( \frac{dy}{dv} = u^v \ln(u) ).
- Substitute ( \frac{du}{dx} ), ( \frac{dv}{dx} ), ( \frac{dy}{du} ), and ( \frac{dy}{dv} ) into the differentiation formula.
- Compute ( \frac{dy}{dx} ) by multiplying and adding the appropriate terms.

So, the derivative ( \frac{dy}{dx} ) of ( y = (\sin(3x))^{\ln(x)} ) is:

[ \frac{dy}{dx} = \frac{du}{dx} vu^{v-1} + \frac{dv}{dx}u^v \ln(u) ]

[ = (3\cos(3x))(\ln(x) \sin(3x)^{\ln(x)-1}) + \frac{1}{x}(\sin(3x))^{\ln(x)} \ln(\sin(3x)) ]

This is the derivative of the given function.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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