Question

If we have a function as $y = \sin [\cos (\sin x)]$ then $\dfrac{{dy}}{{dx}} = $
$1) - \cos [\cos (\sin x)]\sin (\cos x)\cos x$
$2) - \cos [\cos (\sin x)]\sin (\sin x)\cos x$
$3)\cos [\cos (\sin x)]\sin (\cos x)\cos x$
$4)\cos [\cos (\sin x)]\sin (\sin x)\cos x$

Answer 1

Hint: 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 $\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)$
$\dfrac{{d(\sin x)}}{{dy}} = \cos x,\dfrac{{d(\cos x)}}{{dy}} = - \sin x$

Complete step-by-step solution:
Since the given that we have $y = \sin [\cos (\sin x)]$
Now applying the chain rule, we get $\dfrac{{d(\sin [\cos (\sin x)])}}{{dy}} = \cos [\cos (\sin x)].\dfrac{d}{{dx}}[\cos (\sin x)]$ were taking $f = \sin ,g = \cos $ and also $\dfrac{{d(\sin x)}}{{dy}} = \cos x$
Again, applying the chain rule, we get $\dfrac{d}{{dx}}[\cos (\sin x)] = - \sin (\sin x).\dfrac{d}{{dx}}(\sin x)$ where $f = \cos ,g = \sin x$ and $\dfrac{{d(\cos x)}}{{dy}} = - \sin x$
Again, applying the normal derivative for $\dfrac{d}{{dx}}(\sin x)$ then we get $\dfrac{d}{{dx}}(\sin x) = \cos x$
Now combining all we have $\dfrac{d}{{dx}}(\sin [\cos (\sin x)]) = \cos [\cos (\sin x)].( - \sin \sin x\cos x)$
Thus, by the multiplication operation, we get $\dfrac{{dy}}{{dx}} = - \cos [\cos (\sin x)][\sin (\sin x)\cos x]$
Therefore, the option $2) - \cos [\cos (\sin x)]\sin (\sin x)\cos x$ 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 $\dfrac{{\sin }}{{\cos }} = \tan $ and $\tan = \dfrac{1}{{\cot }}$
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 $x$ raised to the power is denoted by $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ .
Differentiation and integration are inverse processes like a derivative of \[\dfrac{{d({x^2})}}{{dx}} = 2x\] and the integration is $\int {2xdx = \dfrac{{2{x^2}}}{2}} \Rightarrow {x^2}$