How do you differentiate $y=2\cos ecx+5\cos x$

Hint: The derivative of the above trigonometric function can be found out in 3 steps. In step1, we find the derivative of the cosine function. In step2, we find the derivative of a cosecant function. In step3, we substitute the derivatives of cosine and cosecant in the function y to get the derivative of the given function.

Complete step-by-step solution:
In the above question, we are supposed to find the derivative of the y. The derivative of the trigonometric function y can be found by using three steps.
Step1:
The derivative of the cosine function:
The derivative of the cosine function is the negative of the sine function.
The derivative of the cosine function is given by,
$\Rightarrow \dfrac{d}{dx}\left( \cos x \right)=\left( -\sin x \right)$
Step2:
The derivative of the cosecant function:
From trigonometry,
We know that$\cos ecx=\dfrac{1}{\sin x}$
$\Rightarrow \dfrac{d}{dx}\left( \cos ecx \right)=\dfrac{d}{dx}\left( \dfrac{1}{\sin x} \right)$
According to the formula of derivatives,
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{\left( \dfrac{d}{dx}\left( u \right)\times v-u\times \dfrac{d}{dx}\left( v \right) \right)}{{{\left( v \right)}^{2}}}$
Where u and v are any differentiable functions.
Following the same,
We get
$\Rightarrow \dfrac{d}{dx}\left( \cos ecx \right)=\dfrac{\left( \dfrac{d}{dx}\left( 1 \right)\times \sin x-1\times \dfrac{d}{dx}\left( \sin x \right) \right)}{{{\left( \sin x \right)}^{2}}}$
The derivative of any constant results in a zero.
$\dfrac{d}{dx}\left( 1 \right)=0$
The derivative of the sine function is the cosine function.
$\dfrac{d}{dx}\left( \sin x \right)=\cos x$
Upon substituting we get,
$\Rightarrow \dfrac{d}{dx}\left( \cos ecx \right)=\dfrac{\left( 0\times \sin x-1\times \cos x \right)}{{{\left( \sin x \right)}^{2}}}$
Now evaluate further.
$\Rightarrow \dfrac{d}{dx}\left( \cos ecx \right)=-\dfrac{\left( 1\times \cos x \right)}{{{\left( \sin x \right)}^{2}}}$
The above expression can be written as
$\Rightarrow \dfrac{d}{dx}\left( \cos ecx \right)=\dfrac{\left( -\cos x \right)}{\sin x}\times \dfrac{1}{\sin x}$
The cotangent function in trigonometry is the ratio of cosine and sine functions.
$\dfrac{\cos x}{\sin x}=\cot x$
The inverse of the sine function is the cosecant function.
$\dfrac{1}{\sin x}=\cos ecx$
Substituting the same,
$\Rightarrow \dfrac{d}{dx}\left( \cos ecx \right)=-\cot x\times \cos ecx$
Step3:
We need to find out the derivative of the function y.
The derivative of the function y is denoted by$\dfrac{dy}{dx}$.
$\Rightarrow y=2\cos ecx+5\cos x$
Differentiating on both sides,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( 2\cos ecx+5\cos x \right)$
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( 2\cos ecx \right)+\dfrac{d}{dx}\left( 5\cos x \right)$
Now bring the constant outside.
$\Rightarrow \dfrac{dy}{dx}=2\times \dfrac{d}{dx}\left( \cos ecx \right)+5\times \dfrac{d}{dx}\left( \cos x \right)$
Substituting the derivatives of the trigonometric functions from step 1 and step 2,
$\Rightarrow \dfrac{dy}{dx}=2\times \left( \left( -\cot x \right)\times \cos ecx \right)+5\times \left( -\sin x \right)$
$\Rightarrow \dfrac{dy}{dx}=-2\cot x\cos ecx-5\sin x$
Hence, the derivative of the given function $y$ is $-2\cot x\cos ecx-5\sin x$.

Note: The derivatives of all trigonometric functions should be known to solve this question easily. The basic relations between the trigonometric functions like$\cot x$ in terms of $\sin x$and $\cos x$ is to be remembered to derive the derivative of the given function.