Question

If \[f(x) = \cos x, g(x) = \log x\] and \[y = \left( {gof} \right)(x)\], then \[\dfrac{{dy}}{{dx}}\] at \[x = 0\] is
A. \[0\]
B. \[1\]
C. \[ - 1\]
D. None of these

Answer 1

Hint: Firstly work on the expression of y using the composition of functions correctly. Differentiate both the sides of the equation of y using the values of derivatives of functions. Substitute the given value of x so as to find the particular solution.
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.

Complete step-by-step answer:
We are given \[f(x) = \cos x\] and \[g(x) = \log x\]
Thus the value of the function \[y = \left( {gof} \right)(x)\] can be written as
\[ = g(\cos x)\]
Now, as \[g(x) = \log x\] we get,
\[ = log(\cos x)\]
Therefore we get \[y = log(\cos x)\]
Differentiating both sides with respect to \[x\] we get ,
\[\dfrac{{dy}}{{dx}} = \left( {\dfrac{1}{{\cos x}}} \right)\left( { - \sin x} \right)\]
\[ = - \tan x\]
Putting \[x = 0\] we get ,
\[\dfrac{{dy}}{{dx}} = - \tan 0 = 0\]
So, the correct answer is “Option A”.

Note: Simply put, differentiable means the derivative exists at every point in its domain. Consequently, the only way for the derivative to exist is if the function also exists (i.e., is continuous) on its domain. Thus, a differentiable function is also a continuous function. In order to find the derivative at a point, substitute the value of x and y in the equation obtained after differentiating the function.