Question

If \[f(x) = \cos x{\rm{ and }}g(x) = \log x\] and \[y = (g \circ f)(x)\] then find the value of \[\dfrac{{dy}}{{dx}}{\rm{ at }}x = 0\].
A.0
B. 1
C. -1
D. none of these

Answer 1

Hint: First substitute f(x) for x in the equation of g(x) to obtain y, then differentiate the function with respect to x by applying chain rule and substitute 0 for x in \[\dfrac{{dy}}{{dx}}\] to obtain the solution.



Formula Used:The chain rule of derivative is,
\[\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(x)\dfrac{d}{{dx}}g(x)\]
And,
\[\dfrac{d}{{dx}}\left[ {\log x} \right] = \dfrac{1}{x}\],
\[\dfrac{d}{{dx}}\left[ {\cos x} \right] = - \sin x\]



Complete step by step solution:It is given that \[f(x) = \cos x{\rm{ and }}g(x) = \log x\]and \[y = (g \circ f)(x)\].
Therefore,
 \[\begin{array}{c}y = g(f(x))\\ = g(\cos x)\\ = \log (\cos x)\end{array}\]
Now, differentiate \[y = \log (\cos x)\] with respect to x,
\[\begin{array}{c}\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {\log (\cos x)} \right]\\ = \dfrac{1}{{\cos x}} \times \left[ { - \sin x} \right]\\ = - \tan x\end{array}\]
Hence,
 \[\begin{array}{c}{\left[ {\dfrac{{dy}}{{dx}}} \right]_{x = 0}} = {\left[ { - \tan x} \right]_{x = 0}}\\ = - \tan 0\\ = 0\end{array}\]



Option ‘A’ is correct

Additional Information:The derivative of a function represents the rate of change of the curve. The derivative of a function at x = a represents the rate of change of the function at x = a. If the derivative at x=a is zero then the function does not change at x = a.



Note: Sometime students get confused with the term \[g(\cos x)\] and unable to do the rest of the calculation. So, \[g(\cos x)\] means that we are replacing x by \[\cos x\] in the given equation of g(x).