find the derivative of y with respect to v.\ny = v\\sqrt{1 - v^{2}}-5\\sin^{-1}v\n\\frac{dy}{dv}=□

Explanation:

Step1: Apply product - rule on $v\sqrt{1 - v^{2}}$

The product - rule states that if $u = v$ and $w=\sqrt{1 - v^{2}}=(1 - v^{2})^{\frac{1}{2}}$, then $(uw)'=u'w + uw'$. We know that $u'=\frac{du}{dv}=1$. To find $w'$, use the chain - rule. Let $t = 1 - v^{2}$, then $w=t^{\frac{1}{2}}$. By the chain - rule $\frac{dw}{dv}=\frac{dw}{dt}\cdot\frac{dt}{dv}$. $\frac{dw}{dt}=\frac{1}{2}t^{-\frac{1}{2}}$ and $\frac{dt}{dv}=-2v$. So $w'=\frac{1}{2}(1 - v^{2})^{-\frac{1}{2}}\cdot(-2v)=\frac{-v}{\sqrt{1 - v^{2}}}$. Then $(v\sqrt{1 - v^{2}})'=1\cdot\sqrt{1 - v^{2}}+v\cdot\frac{-v}{\sqrt{1 - v^{2}}}=\sqrt{1 - v^{2}}-\frac{v^{2}}{\sqrt{1 - v^{2}}}=\frac{1 - v^{2}-v^{2}}{\sqrt{1 - v^{2}}}=\frac{1 - 2v^{2}}{\sqrt{1 - v^{2}}}$.

Step2: Find the derivative of $-5\sin^{-1}v$

The derivative of $\sin^{-1}v$ is $\frac{1}{\sqrt{1 - v^{2}}}$. So the derivative of $-5\sin^{-1}v$ is $-5\cdot\frac{1}{\sqrt{1 - v^{2}}}=-\frac{5}{\sqrt{1 - v^{2}}}$.

Step3: Find $\frac{dy}{dv}$

$\frac{dy}{dv}=(v\sqrt{1 - v^{2}}-5\sin^{-1}v)'=(v\sqrt{1 - v^{2}})'-(5\sin^{-1}v)'$. Substitute the results from Step 1 and Step 2: $\frac{dy}{dv}=\frac{1 - 2v^{2}}{\sqrt{1 - v^{2}}}-\frac{5}{\sqrt{1 - v^{2}}}=\frac{1 - 2v^{2}-5}{\sqrt{1 - v^{2}}}=\frac{-2v^{2}-4}{\sqrt{1 - v^{2}}}$.

Answer:

$\frac{-2v^{2}-4}{\sqrt{1 - v^{2}}}$