Question

Find the derivative of the following:
$y=\cos e{{c}^{2}}x+{{\cot }^{2}}x$

Answer 1

Hint: These are the functions of the form ${{\left[ F(x) \right]}^{n}}$, where $F(x)$ is $\cos ecx$ and $\cot x$ in the first and second term respectively and ‘n’ is 2. Use the formula: \[\dfrac{d}{dx}{{\left[ F\left( x \right) \right]}^{n}}=n{{\left[ F\left( x \right) \right]}^{n-1}}\dfrac{d}{dx}\left[ F\left( x \right) \right]\], to find the derivative of the given trigonometric function. Use the derivative of $\cot x$ equal to $-\cot x\cos ecx$ and the derivative of \[\cot x\] equal to \[-\cos e{{c}^{2}}x\].

Complete step-by-step answer:
We have been provided with the trigonometric function: $y=\cos e{{c}^{2}}x+{{\cot }^{2}}x$. We have to find its derivative.
Clearly they can be seen as the function of the form: ${{\left[ F(x) \right]}^{n}}$. Let us find the derivative of both the terms of ‘y’ separately.
The first term is $\cos e{{c}^{2}}x$. Applying the formula: \[\dfrac{d}{dx}{{\left[ F\left( x \right) \right]}^{n}}=n{{\left[ F\left( x \right) \right]}^{n-1}}\dfrac{d}{dx}\left[ F\left( x \right) \right]\], where $F(x)=\cos ecx$ and n = 2, we get,
\[\begin{align}
  & \dfrac{d}{dx}{{\left[ \cos ecx \right]}^{2}}=2{{\left[ \cos ecx \right]}^{2-1}}\dfrac{d}{dx}\left[ \cos ecx \right] \\
 & =2\cos ecx\dfrac{d}{dx}\left[ \cos ecx \right] \\
\end{align}\]
Using the relation: $\dfrac{d}{dx}\left[ \cos ecx \right]=-\cos ecx\cot x$, we get,
\[\begin{align}
  & \dfrac{d}{dx}{{\left[ \cos ecx \right]}^{2}}=2\cos ecx\left( -\cos ecx\cot x \right) \\
 & =-2\cos e{{c}^{2}}x\cot x \\
\end{align}\]
The second term is ${{\cot }^{2}}x$. Applying the formula: \[\dfrac{d}{dx}{{\left[ g\left( x \right) \right]}^{n}}=n{{\left[ g\left( x \right) \right]}^{n-1}}\dfrac{d}{dx}\left[ g\left( x \right) \right]\], where $g(x)=\cot x$ and n = 2, we get,
\[\begin{align}
  & \dfrac{d}{dx}{{\left[ \cot x \right]}^{2}}=2{{\left[ \cot x \right]}^{2-1}}\dfrac{d}{dx}\left[ \cot x \right] \\
 & =2\cot x\dfrac{d}{dx}\left[ \cot x \right] \\
\end{align}\]
Using the relation: $\dfrac{d}{dx}\left[ \cot x \right]=-\cos e{{c}^{2}}x$, we get,
\[\begin{align}
  & \dfrac{d}{dx}{{\left[ \cot x \right]}^{2}}=2\cot x\left( -\cos e{{c}^{2}}x \right) \\
 & =-2\cot x\cos e{{c}^{2}}x \\
\end{align}\]
Now, we have,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d}{dx}\left[ \cos e{{c}^{2}}x+{{\cot }^{2}}x \right] \\
 & =\dfrac{d}{dx}\left[ \cos e{{c}^{2}}x \right]+\dfrac{d}{dx}\left[ {{\cot }^{2}}x \right] \\
 & =-2\cos e{{c}^{2}}x\cot x+\left( -2\cos e{{c}^{2}}x\cot x \right) \\
 & =-4\cos e{{c}^{2}}x\cot x \\
\end{align}$

Note: One may note that we have an alternate method to solve this question. We have to use the identity: $\cos e{{c}^{2}}x=1+{{\cot }^{2}}x$ to convert the given function into a function containing a single trigonometric ratio. The main advantage of this method is that we have to find the derivative of only one trigonometric ratio, that is either $\cos ecx$ or $\cot x$. The other term will be a constant and its derivative will be zero.