# Find g'(x), where g(x)= \int_{tan x}^{x \arccos x} \frac {dt}{1 +3 \sqrt t}.

## Question:

Find g'(x), where {eq}g(x) {/eq} = {eq}\int_{tan x}^{x \arccos x} \frac {dt}{1 +3 \sqrt t} {/eq}.

## Antiderivatives and the First Fundamental Theorem of Calculus:

Suppose that {eq}f(x) {/eq} and {eq}F(x) {/eq} are continuous functions on a closed, bounded interval {eq}[a,b] {/eq}. If {eq}F'(x)=f(x) {/eq} on the open interval {eq}(a,b) {/eq}, then we say that {eq}F {/eq} is an antiderivative of {eq}f {/eq} on {eq}[a,b] {/eq}.

If {eq}F {/eq} is an antiderivative of {eq}f {/eq} on {eq}[a,b] {/eq}, then the first fundamental theorem of calculus states that

{eq}\displaystyle \int_a^b f(x) \, dx = F(b)-F(a) \, . {/eq}

## Answer and Explanation:

Let {eq}F(t) {/eq} be an antiderivative of the function {eq}\frac{1}{1+3\sqrt{t}} {/eq}. If {eq}\displaystyle g(x)=\int_{\tan x}^{x\arccos x} \frac{dt}{1+3\sqrt{t}} {/eq}, then the first fundamental theorem of calculus gives:

{eq}\displaystyle g(x)=F(x \arccos x) - F(\tan x) \, . {/eq}

Differentiating {eq}g {/eq}, we therefore have

{eq}\begin{align*} g'(x)&=\frac{d}{dx}\left[F(x\arccos x)-F(\tan x)\right]\\ &=\frac{d}{dx}(x\arccos x)F'(x\arccos x)-\frac{d}{dx}(\tan x)F'(\tan x)&&\text{(by the chain rule)}\\ &=\left(\arccos x - \frac{x}{\sqrt{1-x^2}}\right)F'(x\arccos x)-\sec^2 x F'(\tan x)&&\text{(evaluating derivatives)}\\ &=\left(\arccos x - \frac{x}{\sqrt{1-x^2}}\right)\left(\frac{1}{1+3\sqrt{x\arccos x}}\right) - \sec^2 x \left(\frac{1}{1+3\sqrt{\tan x}}\right)&&\text{(since }F\text{ is an antiderivative).} \end{align*} {/eq}

In summary, we've computed that

{eq}\boxed{g'(x)=\left(\arccos x - \frac{x}{\sqrt{1-x^2}}\right)\left(\frac{1}{1+3\sqrt{x\arccos x}}\right) - \sec^2 x \left(\frac{1}{1+3\sqrt{\tan x}}\right)}\,. {/eq}

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