Question

If $y={{\cos }^{-1}}\left( \dfrac{3x+4\sqrt{1-{{x}^{2}}}}{5} \right)$ find $\dfrac{dy}{dx}$ .

Answer 1

Hint: Before differentiating, we need to simplify the expression. We do so by first taking cosine on both sides and then multiplying with $5$ . The expression becomes $5\cos y=3x+4\sqrt{1-{{x}^{2}}}$ . We then differentiate both sides with respect to x and apply the necessary chain rules wherever needed.

Complete step by step answer:
The given equation that we have at our disposal is,
$y={{\cos }^{-1}}\left( \dfrac{3x+4\sqrt{1-{{x}^{2}}}}{5} \right)$
Taking cosine on both sides of the above equation, the above equation thus becomes,
$\Rightarrow \cos y=\cos \left( {{\cos }^{-1}}\left( \dfrac{3x+4\sqrt{1-{{x}^{2}}}}{5} \right) \right)$
We all know the property of cosines and its inverse which is $\cos \left( {{\cos }^{-1}}x \right)=x$ . This analogy gives,
$\Rightarrow \cos y=\dfrac{3x+4\sqrt{1-{{x}^{2}}}}{5}....\left( i \right)$
Multiplying $5$ on both sides of the above equation, the above equation thus becomes,
$\Rightarrow 5\cos y=3x+4\sqrt{1-{{x}^{2}}}$
Differentiating the above equation on both sides with respect to x, the above equation thus becomes,
$\Rightarrow 5\dfrac{d\left( \cos y \right)}{dx}=\dfrac{d}{dx}\left( 3x+4\sqrt{1-{{x}^{2}}} \right)$
The derivative of $\cos x$ is $-\sin x$ . Applying chain rule to the left-hand side of the above equation, the above equation thus becomes,
$\Rightarrow 5\left( -\sin y \right)\times \dfrac{dy}{dx}=\dfrac{d}{dx}\left( 3x+4\sqrt{1-{{x}^{2}}} \right)$
The derivative of x is simply $1$ . So, implementing this in the above equation, the above equation thus becomes,
$\Rightarrow 5\left( -\sin y \right)\times \dfrac{dy}{dx}=3\times 1+\dfrac{d\left( 4\sqrt{1-{{x}^{2}}} \right)}{dx}$
Applying chain rule to the right hand side of the above equation, the above equation thus becomes,
$\Rightarrow 5\left( -\sin y \right)\times \dfrac{dy}{dx}=3+4\times \left( \dfrac{d\left( \sqrt{1-{{x}^{2}}} \right)}{d\left( 1-{{x}^{2}} \right)}\times \dfrac{d\left( 1-{{x}^{2}} \right)}{dx} \right)$
The derivative of $\sqrt{u}$ is $\dfrac{1}{2\sqrt{u}}$ . So, implementing this in the above equation, the above equation thus becomes,
$\Rightarrow 5\left( -\sin y \right)\times \dfrac{dy}{dx}=3+4\times \left( \dfrac{1}{2\sqrt{1-{{x}^{2}}}}\times \dfrac{d\left( 1-{{x}^{2}} \right)}{dx} \right)$
The derivative of any constant is zero and that of ${{x}^{2}}$ is $2x$ . So, implementing these in the above equation, the above equation thus becomes,
$\Rightarrow 5\left( -\sin y \right)\times \dfrac{dy}{dx}=3+4\times \left( \dfrac{1}{2\sqrt{1-{{x}^{2}}}}\times \left( 0-2x \right) \right)$
Simplifying, we get,
$\begin{align}
  & \Rightarrow -5\sin y\times \dfrac{dy}{dx}=3-\dfrac{4x}{\sqrt{1-{{x}^{2}}}} \\
 & \Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{5\sin y}\times \left( 3-\dfrac{4x}{\sqrt{1-{{x}^{2}}}} \right)....\left( ii \right) \\
\end{align}$
From equation i, we get,
\[siny=\sqrt{1-{{\left( \dfrac{3x+4\sqrt{1-{{x}^{2}}}}{5} \right)}^{2}}}\]
Equation ii becomes,
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{5}\times \left( 3-\dfrac{4x}{\sqrt{1-{{x}^{2}}}} \right)\times \dfrac{1}{\sqrt{1-{{\left( \dfrac{3x+4\sqrt{1-{{x}^{2}}}}{5} \right)}^{2}}}} \\
 & \Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{5}\times \left( 3-\dfrac{4x}{\sqrt{1-{{x}^{2}}}} \right)\times \dfrac{1}{\sqrt{1-\left( \dfrac{9{{x}^{2}}+24x\sqrt{1-{{x}^{2}}}+16\left( 1-{{x}^{2}} \right)}{25} \right)}} \\
 & \Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{5}\times \left( 3-\dfrac{4x}{\sqrt{1-{{x}^{2}}}} \right)\times \dfrac{5}{\sqrt{5-9{{x}^{2}}-24x\sqrt{1-{{x}^{2}}}-16+16{{x}^{2}}}} \\
 & \Rightarrow \dfrac{dy}{dx}=-\left( 3-\dfrac{4x}{\sqrt{1-{{x}^{2}}}} \right)\times \dfrac{1}{\sqrt{7{{x}^{2}}-24x\sqrt{1-{{x}^{2}}}-11}} \\
\end{align}$

Therefore, we can conclude that the value of $\dfrac{dy}{dx}$ is $-\left( 3-\dfrac{4x}{\sqrt{1-{{x}^{2}}}} \right)\times \dfrac{1}{\sqrt{7{{x}^{2}}-24x\sqrt{1-{{x}^{2}}}-11}}$ .

Note: From the long and tedious derivation, it is very clear to us that the problem requires tremendous attention and a small mistake anywhere can cause a chain reaction of errors. Students often try to differentiate the expression right from the beginning, which is technically correct but becomes excessively long and more prone to mistakes. So, simplifying the given expression at the beginning is required.