If $$ H(x)=\int_{-1}^{\cos ^{-1} x} an \left(\frac{1}{4} Q ight) d Q $$, find $$ H^{\prime}(x) $$ and $$ H^{\prime}\left(-\frac{1}{2} ight) $$

Question

If $$ H(x)=\int_{-1}^{\cos ^{-1} x} 	an \left(\frac{1}{4} Qight) d Q $$, find $$ H^{\prime}(x) $$ and $$ H^{\prime}\left(-\frac{1}{2}ight) $$

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To find $$ H'(x) $$, we will use the Fundamental Theorem of Calculus and the chain rule. The function $$ H(x) $$ is defined as:

$$
H(x) = \int_{-1}^{\cos^{-1} x} \tan\left(\frac{1}{4} Q\right) dQ
$$

To differentiate $$ H(x) $$ with respect to $$ x $$, we apply the Fundamental Theorem of Calculus, which states that if $$ F(Q) $$ is an antiderivative of $$ f(Q) $$, then:

$$
\frac{d}{dx} \int_{a}^{g(x)} f(Q) \, dQ = f(g(x)) \cdot g'(x)
$$

In our case, $$ f(Q) = \tan\left(\frac{1}{4} Q\right) $$ and $$ g(x) = \cos^{-1} x $$. Thus, we have:

$$
H'(x) = \tan\left(\frac{1}{4} g(x)\right) \cdot g'(x)
$$

First, we need to compute $$ g'(x) $$:

$$
g(x) = \cos^{-1} x \implies g'(x) = -\frac{1}{\sqrt{1 - x^2}}
$$

Now substituting $$ g(x) $$ into $$ H'(x) $$:

$$
H'(x) = \tan\left(\frac{1}{4} \cos^{-1} x\right) \cdot \left(-\frac{1}{\sqrt{1 - x^2}}\right)
$$

Thus, we have:

$$
H'(x) = -\frac{\tan\left(\frac{1}{4} \cos^{-1} x\right)}{\sqrt{1 - x^2}}
$$

Next, we need to evaluate $$ H'\left(-\frac{1}{2}\right) $$. First, we compute $$ \cos^{-1}\left(-\frac{1}{2}\right) $$:

$$
\cos^{-1}\left(-\frac{1}{2}\right) = \frac{2\pi}{3}
$$

Now substituting $$ x = -\frac{1}{2} $$ into $$ H'(x) $$:

$$
H'\left(-\frac{1}{2}\right) = -\frac{\tan\left(\frac{1}{4} \cdot \frac{2\pi}{3}\right)}{\sqrt{1 - \left(-\frac{1}{2}\right)^2}}
$$

Calculating $$ \sqrt{1 - \left(-\frac{1}{2}\right)^2} $$:

$$
\sqrt{1 - \left(-\frac{1}{2}\right)^2} = \sqrt{1 - \frac{1}{4}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}
$$

Now we compute $$ \tan\left(\frac{1}{4} \cdot \frac{2\pi}{3}\right) $$:

$$
\frac{1}{4} \cdot \frac{2\pi}{3} = \frac{\pi}{6} \implies \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}
$$

Putting it all together:

$$
H'\left(-\frac{1}{2}\right) = -\frac{\frac{1}{\sqrt{3}}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} \cdot \frac{2}{\sqrt{3}} = -\frac{2}{3}
$$

Thus, the final result is:

$$
H'\left(-\frac{1}{2}\right) = -\frac{2}{3}
$$
$$ H'(x) = -\frac{\tan\left(\frac{1}{4} \cos^{-1} x\right)}{\sqrt{1 - x^2}} $$ $$ H'\left(-\frac{1}{2}\right) = -\frac{2}{3} $$

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