# How do you differentiate #g(a)= sin(arcsin(5a))#?

#(dg(a))/(da) = 5#

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To differentiate g(a) = sin(arcsin(5a)):

- Recognize that arcsin(5a) is the inverse function of sin(x), which means arcsin(5a) yields an angle whose sine is 5a.
- Differentiate arcsin(5a) with respect to 'a' using the chain rule.
- Then differentiate sin(arcsin(5a)) using the chain rule.

The derivative of g(a) with respect to 'a' is:

g'(a) = (1 / √(1 - (5a)^2)) * 5*cos(arcsin(5a))

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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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