Question

Find $\dfrac{{dy}}{{dx}}$ if $y = \cos (1 - x)$

Answer 1

Here we are asked to find the derivative of the given expression. As we can see that the given expression is in trigonometric functions. We can use the standard formula to find the derivative of the trigonometric functions which has been given in the formula section. Also, in the given trigonometric function angle is also an expression these types of functions can be differentiated by using chain rule that is we will first differentiate the outer function then multiply it by the differentiation of the inner function.
Formula used:
Chain rule of differentiation $\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)$
$\dfrac{d}{{dx}}\cos x = - \sin x$

Complete step by step answer:
Since given that $y = \cos (1 - x)$ and we need to find its derivative value with respect to the variable $x$
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.
Now applying the chain rule, we have, $\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)$and note that $\dfrac{d}{{dx}}\cos x = - \sin x$ apply this in the given function $y = \cos (1 - x)$ then we get $\dfrac{{dy}}{{dx}} = - \sin (1 - x)\dfrac{d}{{dx}}(1 - x)$ because since we know that ${f^1}(g(x)) = \dfrac{d}{{dx}}(\cos (1 - x)) \Rightarrow - \sin (1 - x)$ and also ${g^1}(x) = \dfrac{d}{{dx}}(1 - x)$
Further solving we get ${g^1}(x) = \dfrac{d}{{dx}}(1 - x) \Rightarrow - 1$
Hence substituting the values in the above we get $\dfrac{{dy}}{{dx}} = - \sin (1 - x)\dfrac{d}{{dx}}(1 - x) \Rightarrow \dfrac{{dy}}{{dx}} = - \sin (1 - x)( - 1)$
Thus we get the derivative as $\dfrac{{dy}}{{dx}} = \sin (1 - x)$ which is the required answer.

Note:
The main concept used in the problem is the chain rule. We also must know the derivatives of the basic functions like tangent and sec.
We 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 $\tan = \dfrac{1}{{\cot }}$
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 $\int {2xdx = \dfrac{{2{x^2}}}{2}} \Rightarrow {x^2}$