Question

How do you find the derivative of \[f\left( x \right)=cos\left( \sin \left( 4x \right) \right)\]?

Answer 1

Hint: Assume \[p\left( x \right)=4x,g\left( x \right)=\sin x\] and \[h\left( x \right)=\cos x\] and write the given function as a composite function: \[y=h\left[ g\left( p\left( x \right) \right) \right]\]. Now, differentiate both sides of the function with respect to the variable x and use the chain rule of differentiation to find the derivative of \[h\left[ g\left( p\left( x \right) \right) \right]\]. Use the relation: - \[\dfrac{d\left[ h\left( g\left( p\left( x \right) \right) \right) \right]}{dx}=h'\left( g\left( p\left( x \right) \right) \right)\times g'\left( p\left( x \right) \right)\times p'\left( x \right)\] to get the answer. Use the basic formulas:
$\dfrac{d\sin x}{dx}=\cos x,\dfrac{d\cos x}{dx}=-sinx$.

Complete step by step solution:
Here, we have been provided with the function \[f\left( x \right)=cos\left( \sin \left( 4x \right) \right)\] and we are asked to find its derivative.
We can convert the given function into a composite function because we have a combination of several functions. So, assuming \[p\left( x \right)=4x,g\left( x \right)=\sin x\] and \[h\left( x \right)=\cos x\], we have,
\[\Rightarrow f\left( x \right)=h\left[ g\left( p\left( x \right) \right) \right]\]
Differentiating both the sides with respect to the variable x, we have,
\[\begin{align}
  & \Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=\dfrac{d\left[ h\left( g\left( p\left( x \right) \right) \right) \right]}{dx} \\
 & \Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=h'\left[ g\left( p\left( x \right) \right) \right]\times g'\left[ p\left( x \right) \right]\times p'\left( x \right) \\
\end{align}\]
Here, we have applied the chain rule of differentiation in the R.H.S. and that is how the derivative of a composite function is determined. What we have to do in the next step is we have to find the derivative of \[h\left[ g\left( p\left( x \right) \right) \right]\] with respect to \[g\left( p\left( x \right) \right)\], the derivative of \[g\left( p\left( x \right) \right)\] with respect to \[p\left( x \right)\], the derivative of \[p\left( x \right)\] with respect to x and then take the product of these three derivatives obtained. So, mathematically we have,
\[\Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=\dfrac{d\left[ h\left( g\left( p\left( x \right) \right) \right) \right]}{d\left[ g\left( p\left( x \right) \right) \right]}\times \dfrac{d\left[ g\left( p\left( x \right) \right) \right]}{d\left[ p\left( x \right) \right]}\times \dfrac{d\left[ p\left( x \right) \right]}{dx}\]
Substituting the assumed values, we get,
\[\Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=\dfrac{d\left[ \cos \left( \sin \left( 4x \right) \right) \right]}{d\left( sin\left( 4x \right) \right)}\times \dfrac{d\left( \sin \left( 4x \right) \right)}{d\left( 4x \right)}\times \dfrac{d\left( 4x \right)}{dx}\]
Using the basic formulas: $\dfrac{d\sin x}{dx}=\cos x,\dfrac{d\cos x}{dx}=-sinx$, we get,
\[\begin{align}
  & \Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=-sin\left( \sin \left( 4x \right) \right)\times \cos \left( 4x \right)\times 4 \\
 & \Rightarrow \dfrac{d\left[ f\left( x \right) \right]}{dx}=-4sin\left( \sin \left( 4x \right) \right)\times \cos \left( 4x \right) \\
\end{align}\]

Hence, the derivative of the given function is: \[-4sin\left( \sin \left( 4x \right) \right)\times \cos \left( 4x \right)\].

Note: One must remember the derivatives of basic functions like: - trigonometric functions, logarithmic functions, exponential functions etc. Here, in the above question we do not have any other method to solve the question. The given function is a composite function we use the chain rule for finding the derivative of such functions. You must remember all the basic rules of differentiation like: - the product rule, chain rule, \[\dfrac{u}{v}\] rule etc. as they are used everywhere in calculus.