Question

find the derivative of y with respect to u.
y = u\sqrt{1 - u^{2}}-\sin^{-1}u\\
\frac{dy}{du}=\square

Explanation:

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

The product - rule states that if $y = f(u)g(u)$, then $y^\prime=f^\prime(u)g(u)+f(u)g^\prime(u)$. Let $f(u)=u$ and $g(u)=\sqrt{1 - u^{2}}=(1 - u^{2})^{\frac{1}{2}}$. First, $f^\prime(u) = 1$. For $g(u)=(1 - u^{2})^{\frac{1}{2}}$, using the chain - rule, if $y = h(k(u))$, then $y^\prime=h^\prime(k(u))k^\prime(u)$. Let $k(u)=1 - u^{2}$ and $h(k)=k^{\frac{1}{2}}$. Then $k^\prime(u)=-2u$ and $h^\prime(k)=\frac{1}{2}k^{-\frac{1}{2}}$. So $g^\prime(u)=\frac{1}{2}(1 - u^{2})^{-\frac{1}{2}}\times(-2u)=\frac{-u}{\sqrt{1 - u^{2}}}$. Then the derivative of $u\sqrt{1 - u^{2}}$ is $1\times\sqrt{1 - u^{2}}+u\times\frac{-u}{\sqrt{1 - u^{2}}}=\sqrt{1 - u^{2}}-\frac{u^{2}}{\sqrt{1 - u^{2}}}=\frac{1 - u^{2}-u^{2}}{\sqrt{1 - u^{2}}}=\frac{1 - 2u^{2}}{\sqrt{1 - u^{2}}}$.

Step2: Apply derivative formula for $\sin^{-1}u$

The derivative of $\sin^{-1}u$ with respect to $u$ is $\frac{1}{\sqrt{1 - u^{2}}}$.

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

$\frac{dy}{du}=\frac{1 - 2u^{2}}{\sqrt{1 - u^{2}}}-\frac{1}{\sqrt{1 - u^{2}}}=\frac{1 - 2u^{2}-1}{\sqrt{1 - u^{2}}}=\frac{-2u^{2}}{\sqrt{1 - u^{2}}}$

Answer:

$\frac{-2u^{2}}{\sqrt{1 - u^{2}}}$