# How do you find the derivative of #y=sqrt(3^x-2^x)#?

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To find the derivative of (y = \sqrt{3^x - 2^x}), you can use the chain rule and the exponential function derivative rules. Here's the step-by-step process:

Given function: [y = \sqrt{3^x - 2^x}]

Rewrite the function to make differentiation clearer: [y = (3^x - 2^x)^{\frac{1}{2}}]

Now, let's differentiate using the chain rule. The chain rule states that if you have a composite function (f(g(x))), then its derivative is (f'(g(x)) \cdot g'(x)). Here, (f(u) = u^{\frac{1}{2}}) and (g(x) = 3^x - 2^x), where (u = g(x)).

First, differentiate (f(u) = u^{\frac{1}{2}}) with respect to (u): [f'(u) = \frac{1}{2}u^{-\frac{1}{2}}]

Then, differentiate (g(x) = 3^x - 2^x) with respect to (x):

- The derivative of (3^x) is (3^x \ln(3)).
- The derivative of (2^x) is (2^x \ln(2)).

So, (g'(x) = 3^x \ln(3) - 2^x \ln(2)).

Now, apply the chain rule by multiplying (f'(g(x))) by (g'(x)): [y' = \frac{1}{2}(3^x - 2^x)^{-\frac{1}{2}} \cdot (3^x \ln(3) - 2^x \ln(2))]

This can be further simplified or left as is, depending on the requirements. So, the derivative of (y = \sqrt{3^x - 2^x}) is: [y' = \frac{3^x \ln(3) - 2^x \ln(2)}{2\sqrt{3^x - 2^x}}]

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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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