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Question: If we have a function as \(y = \sin [\cos (\sin x)]\) then \(\dfrac{{dy}}{{dx}} = \) \(1) - \cos [...

If we have a function as y=sin⁡[cos⁡(sin⁡x)]y = \sin [\cos (\sin x)] then dydx=\dfrac{{dy}}{{dx}} =
1)−cos⁡[cos⁡(sin⁡x)]sin⁡(cos⁡x)cos⁡x1) - \cos [\cos (\sin x)]\sin (\cos x)\cos x
2)−cos⁡[cos⁡(sin⁡x)]sin⁡(sin⁡x)cos⁡x2) - \cos [\cos (\sin x)]\sin (\sin x)\cos x
3)cos⁡[cos⁡(sin⁡x)]sin⁡(cos⁡x)cos⁡x3)\cos [\cos (\sin x)]\sin (\cos x)\cos x
4)cos⁡[cos⁡(sin⁡x)]sin⁡(sin⁡x)cos⁡x4)\cos [\cos (\sin x)]\sin (\sin x)\cos x

Explanation

Solution

First, we need to analyze the given information which is in the trigonometric form.

We can equate the given expression into some form and then we can differentiate using the derivatives of basic functions and applying the chain rule of differentiation.
The trigonometric functions are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
We just need to know the chain rule to solve this problem, which is given below.
Formula used:
Chain rule of differentiation ddx(f(g(x))=f1(g(x))×g1(x)\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)
d(sin⁡x)dy=cos⁡x,d(cos⁡x)dy=−sin⁡x\dfrac{{d(\sin x)}}{{dy}} = \cos x,\dfrac{{d(\cos x)}}{{dy}} = - \sin x

Complete step-by-step solution:
Since the given that we have y=sin⁡[cos⁡(sin⁡x)]y = \sin [\cos (\sin x)]
Now applying the chain rule, we get d(sin⁡[cos⁡(sin⁡x)])dy=cos⁡[cos⁡(sin⁡x)].ddx[cos⁡(sin⁡x)]\dfrac{{d(\sin [\cos (\sin x)])}}{{dy}} = \cos [\cos (\sin x)].\dfrac{d}{{dx}}[\cos (\sin x)] were taking f=sin⁡,g=cos⁡f = \sin ,g = \cos and also d(sin⁡x)dy=cos⁡x\dfrac{{d(\sin x)}}{{dy}} = \cos x
Again, applying the chain rule, we get ddx[cos⁡(sin⁡x)]=−sin⁡(sin⁡x).ddx(sin⁡x)\dfrac{d}{{dx}}[\cos (\sin x)] = - \sin (\sin x).\dfrac{d}{{dx}}(\sin x) where f=cos⁡,g=sin⁡xf = \cos ,g = \sin x and d(cos⁡x)dy=−sin⁡x\dfrac{{d(\cos x)}}{{dy}} = - \sin x
Again, applying the normal derivative for ddx(sin⁡x)\dfrac{d}{{dx}}(\sin x) then we get ddx(sin⁡x)=cos⁡x\dfrac{d}{{dx}}(\sin x) = \cos x
Now combining all we have ddx(sin⁡[cos⁡(sin⁡x)])=cos⁡[cos⁡(sin⁡x)].(−sin⁡sin⁡xcos⁡x)\dfrac{d}{{dx}}(\sin [\cos (\sin x)]) = \cos [\cos (\sin x)].( - \sin \sin x\cos x)
Thus, by the multiplication operation, we get dydx=−cos⁡[cos⁡(sin⁡x)][sin⁡(sin⁡x)cos⁡x]\dfrac{{dy}}{{dx}} = - \cos [\cos (\sin x)][\sin (\sin x)\cos x]
Therefore, the option 2)−cos⁡[cos⁡(sin⁡x)]sin⁡(sin⁡x)cos⁡x2) - \cos [\cos (\sin x)]\sin (\sin x)\cos x is correct.

Note: The main concept used in the given problem is the chain rule. We must know the derivatives of the basic functions like sine and cosine.
We also use simple algebra to simplify the expression that we will get after derivation.
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like sin⁡cos⁡=tan⁡\dfrac{{\sin }}{{\cos }} = \tan and tan⁡=1cot⁡\tan = \dfrac{1}{{\cot }}
Differentiation can be defined as the derivative of the independent variable value and can be used to calculate features in an independence variance per unit modification.
In differentiation, the derivative of xx raised to the power is denoted by ddx(xn)=nxn−1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} .
Differentiation and integration are inverse processes like a derivative of d(x2)dx=2x\dfrac{{d({x^2})}}{{dx}} = 2x and the integration is ∫2xdx=2x22⇒x2\int {2xdx = \dfrac{{2{x^2}}}{2}} \Rightarrow {x^2}