find the derivative of the function.\nh(t) = 7 cot^{-1}(t)+7 cot^{-1}(\\frac{1}{t})\nh(t) =\n\nresources\nrea…

Explanation:

Step1: Recall derivative of inverse - cotangent function

The derivative of $\cot^{-1}(u)$ with respect to $u$ is $-\frac{1}{1 + u^{2}}$. For the function $h(t)=7\cot^{-1}(t)+7\cot^{-1}(\frac{1}{t})$, by the sum - rule of differentiation $(f + g)'=f'+g'$, we have $h'(t)=7\frac{d}{dt}\cot^{-1}(t)+7\frac{d}{dt}\cot^{-1}(\frac{1}{t})$. The derivative of $7\cot^{-1}(t)$ with respect to $t$ is $7\times(-\frac{1}{1 + t^{2}})=-\frac{7}{1 + t^{2}}$ by the constant - multiple rule and the derivative formula for $\cot^{-1}(u)$.

Step2: Use chain - rule for $\cot^{-1}(\frac{1}{t})$

Let $u = \frac{1}{t}=t^{-1}$. Then $\frac{d}{dt}\cot^{-1}(u)=-\frac{1}{1 + u^{2}}\times\frac{du}{dt}$. We know that $\frac{du}{dt}=-t^{-2}=-\frac{1}{t^{2}}$. Substituting $u = \frac{1}{t}$ into $-\frac{1}{1 + u^{2}}\times\frac{du}{dt}$, we get $-\frac{1}{1+\left(\frac{1}{t}\right)^{2}}\times\left(-\frac{1}{t^{2}}\right)$. Simplify $-\frac{1}{1+\frac{1}{t^{2}}}\times\left(-\frac{1}{t^{2}}\right)=-\frac{t^{2}}{t^{2}+1}\times\left(-\frac{1}{t^{2}}\right)=\frac{1}{t^{2}+1}$. So $7\frac{d}{dt}\cot^{-1}(\frac{1}{t})=7\times\frac{1}{t^{2}+1}=\frac{7}{t^{2}+1}$.

Step3: Calculate $h'(t)$

$h'(t)=-\frac{7}{1 + t^{2}}+\frac{7}{t^{2}+1}=0$.

For the function $y = \arctan\left(\sqrt{\frac{1 - x}{1 + x}}\right)$:

Step1: Let $u=\sqrt{\frac{1 - x}{1 + x}}$

By the chain - rule, $\frac{dy}{dx}=\frac{1}{1 + u^{2}}\times\frac{du}{dx}$. First, we find $\frac{du}{dx}$ using the quotient - rule and the chain - rule. If $u=\left(\frac{1 - x}{1 + x}\right)^{\frac{1}{2}}$, by the chain - rule, $\frac{du}{dx}=\frac{1}{2}\left(\frac{1 - x}{1 + x}\right)^{-\frac{1}{2}}\times\frac{d}{dx}\left(\frac{1 - x}{1 + x}\right)$.

Step2: Use quotient - rule for $\frac{d}{dx}\left(\frac{1 - x}{1 + x}\right)$

The quotient - rule states that if $y=\frac{f(x)}{g(x)}$, then $y'=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^{2}}$. Here, $f(x)=1 - x$, $f'(x)=-1$, $g(x)=1 + x$, $g'(x)=1$. So $\frac{d}{dx}\left(\frac{1 - x}{1 + x}\right)=\frac{-1\times(1 + x)-(1 - x)\times1}{(1 + x)^{2}}=\frac{-1 - x - 1 + x}{(1 + x)^{2}}=-\frac{2}{(1 + x)^{2}}$.

Step3: Calculate $\frac{du}{dx}$

$\frac{du}{dx}=\frac{1}{2}\left(\frac{1 + x}{1 - x}\right)^{\frac{1}{2}}\times\left(-\frac{2}{(1 + x)^{2}}\right)=-\frac{1}{\sqrt{(1 - x)(1 + x)^{3}}}$.

Step4: Calculate $\frac{dy}{dx}$

Since $u=\sqrt{\frac{1 - x}{1 + x}}$, then $1 + u^{2}=1+\frac{1 - x}{1 + x}=\frac{1 + x+1 - x}{1 + x}=\frac{2}{1 + x}$. And $\frac{dy}{dx}=\frac{1}{1 + u^{2}}\times\frac{du}{dx}=\frac{1 + x}{2}\times\left(-\frac{1}{\sqrt{(1 - x)(1 + x)^{3}}}\right)=-\frac{1}{2\sqrt{1 - x^{2}}}$.

Answer:

$h'(t)=0$ $y'=-\frac{1}{2\sqrt{1 - x^{2}}}$