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How do you find the derivative of #G(x)=int (tan(t^2))dt# from #[1,x^2]#?

Answer 1

Christopher Allers

#G(x)=int tant^2 \ dt qquad [1,x^2]#

# "ie " G(x)=int_1^(x^2) tant^2 \ dt #

From the FTC:

# G'(x)= tan((x^2)^2) (x^2)' = 2x \ tan x^4 #

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Answer 2

William Abdalla

To find the derivative of (G(x) = \int_{1}^{x^2} \tan(t^2) , dt), you can apply the Fundamental Theorem of Calculus and the Chain Rule.

First, you need to find the antiderivative of (\tan(t^2)). Let (F(t)) be such that (F'(t) = \tan(t^2)). Then, by the Fundamental Theorem of Calculus:

[G(x) = F(x^2) - F(1).]

To find (G'(x)), differentiate (G(x)) with respect to (x):

[G'(x) = \frac{d}{dx} [F(x^2) - F(1)].]

Using the Chain Rule, this becomes:

[G'(x) = F'(x^2) \cdot 2x.]

Substituting (F'(t) = \tan(t^2)), we get:

[G'(x) = \tan(x^4) \cdot 2x.]

Therefore, the derivative of (G(x) = \int_{1}^{x^2} \tan(t^2) , dt) is (G'(x) = 2x \tan(x^4)).

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Answer from HIX Tutor

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.