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Find the derivative of the function.g(x) = \int_4^{x^2}4\sqrt{1+t^2} dt

Question:

Find the derivative of the function.

{eq}g(x) = \int_4^{x^2}4\sqrt{1+t^2} dt {/eq}

Fundamental Theorem of Calculus 1:

If f is continuous on {eq}{/eq} then the function F defined by {eq}{/eq}

is continuous on {eq}{/eq} differentiable on (c,d) and F'(x) = f(x) for each {eq}x \in (c,d) {/eq}. That is, {eq}\displaystyle \dfrac{d}{dx} \left[ \int_a^x f(t) \, dt \right] = f(x). {/eq}

Applying the Chain Rule, we can show that

{eq}\displaystyle \dfrac{d}{dx} \left[ \int_a^{g(x)} f(t) \, dt \right] = f(g(x))\cdot g'(x). {/eq}

Answer and Explanation:

Note that {eq}4\sqrt{1+t^2} {/eq} is continuous on the domain, which is {eq}\mathbb R, {/eq} so Fundamental Theorem of Calculus can be used.

Applying the Chain Rule and Fundamental Theorem of Calculus 1, we get:

{eq}\begin{align*} \displaystyle \dfrac{d}{dx} (g(x)) & = \displaystyle \dfrac{d}{dx}\left( \displaystyle \int_4^{x^2}4\sqrt{1+t^2} dt\right) \\& = 4\sqrt{1+x^2} \cdot \displaystyle \dfrac{d}{dx} (x^2) \\& = 4\sqrt{1+x^2} \cdot 2x \\& = 8x \sqrt{1+x^2} \end{align*} {/eq}


Learn more about this topic:

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The Fundamental Theorem of Calculus

from Math 104: Calculus

Chapter 12 / Lesson 10
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