Question

: If \[y = {\cos ^{ - 1}}(\cos x)\] then \[\dfrac{{dy}}{{dx}}\] is
A.\[1\] In the whole plane
B.\[ - 1\] In the whole plane
C.\[1\] in the 2nd and 3rd quadrants of the plane
D.\[ - 1\] in the 3rd and 4th quadrants of the plane

Answer 1

Hint: In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). Derivatives are a fundamental tool of calculus .Derivatives can be generalized to functions of several real variables. The process of finding a derivative is called differentiation. The reverse process is called anti differentiation.

Complete step-by-step answer:
The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. The tangent line is the best linear approximation of the function near that input value. For this reason, the derivative is often described as the "instantaneous rate of change", the ratio of the instantaneous change in the dependent variable to that of the independent variable.
Let f be a function that has a derivative at every point in its domain. We can then define a function that maps every point \[x\] to the value of the derivative of f at \[x\] . This function is written f′ and is called the derivative function or the derivative of f.
We are given \[y = {\cos ^{ - 1}}(\cos x)\]
Let \[f = {\cos ^{ - 1}}u\] and \[u = \cos x\]
Therefore \[\dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}(\cos x)} \right] = \dfrac{d}{{du}}\left( {{{\cos }^{ - 1}}u} \right)\dfrac{d}{{dx}}\left( {\cos x} \right)\]
\[ = \left( { - \dfrac{1}{{\sqrt {1 - {u^2}} }}} \right)\left( { - \sin x} \right)\]
Substituting value of \[u\] we get ,
\[\dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}(\cos x)} \right] = \left( { - \dfrac{1}{{\sqrt {1 - {{\cos }^2}x} }}} \right)\left( { - \sin x} \right)\]
Therefore we get ,
\[\dfrac{d}{{dx}}\left[ {{{\cos }^{ - 1}}(\cos x)} \right] = \dfrac{{\sin x}}{{\sqrt {{{\sin }^2}x} }} = \pm 1\]
\[\dfrac{{dy}}{{dx}} = 1\] for \[x \in [0,\pi ]\] and \[\dfrac{{dy}}{{dx}} = - 1\] for \[x \in [\pi ,2\pi ]\]
Therefore option (4) is the correct answer.
So, the correct answer is “Option B”.

Note: : Derivatives are a fundamental tool of calculus. Keep in mind all the formulas of differentiation . Derivatives can be generalized to functions of several real variables .Note that integration is known as antidifferentiation.