Question

How do you find the derivative of $\sin \left( cos\left( \tan x \right) \right)$?

Answer 1

Assume $f\left( x \right)=\tan x,g\left( x \right)=\cos x$ and $h\left( x \right)=\sin x$ and write the given function as a composite function: $y=h\left[ g\left( f\left( x \right) \right) \right]$. 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 $h\left[ g\left( f\left( x \right) \right) \right]$. Use the relation: - $\dfrac{d\left[ h\left( g\left( f\left( x \right) \right) \right) \right]}{dx}=h'\left( g\left( f\left( x \right) \right) \right)\times g'\left( f\left( x \right) \right)\times f'\left( x \right)$ to get the answer. Use the basic formulas: $\dfrac{d\sin x}{dx}=\cos x,\dfrac{d\cos x}{dx}=-sinx$ and $\dfrac{d\tan x}{dx}=se{{c}^{2}}x$.

Complete step by step solution:
Here, we have been provided with the function $\sin \left( cos\left( \tan x \right) \right)$ and we are asked to find its derivative.
Now, let us assume this given function as y, that means we have to find the value of $\dfrac{dy}{dx}$. So, we have,
$\Rightarrow y=\sin \left( cos\left( \tan x \right) \right)$
We can convert the given function into a composite function because we have a combination of several functions. So, assuming $f\left( x \right)=\tan x,g\left( x \right)=\cos x$ and $h\left( x \right)=\sin x$, we have,
$\Rightarrow y=h\left[ g\left( f\left( x \right) \right) \right]$
Differentiating both the sides with respect to the variable x, we have,

& \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ h\left( g\left( f\left( x \right) \right) \right) \right]}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=h'\left[ g\left( f\left( x \right) \right) \right]\times g'\left[ f\left( x \right) \right]\times f'\left( x \right) \\\ \end{aligned}$$ Here, we have applied the chain rule of differentiation in the R.H.S. and that is how the derivative of a composite function is determined. What we have to do in the next step is we have to find the derivative of $$h\left[ g\left( f\left( x \right) \right) \right]$$ with respect to $$g\left( f\left( x \right) \right)$$, the derivative of $$g\left( f\left( x \right) \right)$$ with respect to $$f\left( x \right)$$, the derivative of $$f\left( x \right)$$ with respect to x and then take the product of these three derivatives obtained. So, mathematically we have, $$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ h\left( g\left( f\left( x \right) \right) \right) \right]}{d\left[ g\left( f\left( x \right) \right) \right]}\times \dfrac{d\left[ g\left( f\left( x \right) \right) \right]}{d\left[ f\left( x \right) \right]}\times \dfrac{d\left[ f\left( x \right) \right]}{dx}$$ Substituting the assumed values, we get, $$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ \sin \left( \cos \left( \tan x \right) \right) \right]}{d\left( \cos \left( \tan x \right) \right)}\times \dfrac{d\left( \cos \left( \tan x \right) \right)}{d\left( \tan x \right)}\times \dfrac{d\left( \tan x \right)}{dx}$$ Using the basic formulas: $\dfrac{d\sin x}{dx}=\cos x,\dfrac{d\cos x}{dx}=-sinx$ and $\dfrac{d\tan x}{dx}=se{{c}^{2}}x$, we get, $$\begin{aligned} & \Rightarrow \dfrac{dy}{dx}=\cos \left( \cos \left( \tan x \right) \right)\times \left( -\sin \left( \tan x \right) \right)\times {{\sec }^{2}}x \\\ & \Rightarrow \dfrac{dy}{dx}=-\cos \left( \cos \left( \tan x \right) \right)\times \left( \sin \left( \tan x \right) \right)\times {{\sec }^{2}}x \\\ \end{aligned}$$ Hence, the derivative of the given function is: $$-\cos \left( \cos \left( \tan x \right) \right)\times \left( \sin \left( \tan x \right) \right)\times {{\sec }^{2}}x$$. **Note:** One must remember the derivatives of basic functions like: - trigonometric functions, logarithmic functions, exponential functions etc. Here, in the above question we do not have any other method to solve the question. The given function must be converted into a composite function and then the chain rule must be applied. You must remember all the basic rules of differentiation like: - the product rule, chain rule, $$\dfrac{u}{v}$$ rule etc. as they are used everywhere in calculus.