   Question Answers

# Differentiate the given function with respect to x.$\cos (a\cos x + b\sin x)$ for some constants and a and b.  Verified
128.7k+ views
Hint: Here we simply apply chain rule of differentiation and differentiate the function. We will find the differentiation of the term inside the bracket separately and then use it in chain rule.
* Chain rule which will be used within the differentiation is given as $\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)).g'(x)$ where $f'$denotes differentiation of function f with respect to x and $g'$denotes differentiation of function g with respect to x.
* $\dfrac{d}{{dx}}\sin x = \cos x,\dfrac{d}{{dx}}\cos x = - \sin x$

We have the function as $\cos (a\cos x + b\sin x)$
Differentiation of the function will be written as
$\dfrac{d}{{dx}}\cos (a\cos x + b\sin x)$ … (1)
which is of the form $\dfrac{d}{{dx}}\left[ {f(g(x))} \right]$
Here we apply the chain rule of differentiation to solve the differentiation. Chain rule gives us
$\dfrac{d}{{dx}}\left[ {f(g(x))} \right] = f'(g(x)).g'(x)$
Here, $f(g(x)) = \cos (a\cos x + b\sin x),g(x) = (a\cos x + b\sin x)$
$f'(x) = \dfrac{{df(x)}}{{dx}} = \dfrac{{d\cos (g(x))}}{{dx}} = - \sin (g(x))$ and $g'(x) = \dfrac{{dg(x)}}{{dx}} = \dfrac{{d(a\cos x + b\sin x)}}{{dx}} = - a\sin x + b\cos x$
Substituting the values in the equation
$\Rightarrow \dfrac{d}{{dx}}\left[ {\cos (a\cos x + b\sin x} \right] = - \sin (g(x)) \times ( - a\sin x + b\cos x)$
Substitute the value of $g(x) = a\cos x + b\sin x$.
$\Rightarrow \dfrac{d}{{dx}}\left[ {\cos (a\cos x + b\sin x} \right] = - \sin (a\cos x + b\sin x) \times ( - a\sin x + b\cos x)$
Simplifying the equation we get
$\Rightarrow \dfrac{d}{{dx}}\left[ {\cos (a\cos x + b\sin x} \right] = (a\sin x - b\cos x).\sin (a\cos x + b\sin x)$
So, the differentiation of $\cos (a\cos x + b\sin x)$ is $(a\sin x - b\cos x).\sin (a\cos x + b\sin x)$.

Note: Students are very likely to make mistake in the applying of chain rule to the question as they might take differentiation of $f(g(x))$ and try to write differentiation of the term within the bracket along and then again multiply with differentiation of $g(x)$ which is wrong, when we are finding differentiation of $f(g(x))$ we don’t consider the term inside the bracket, we just differentiate the function that is outside and then we take differentiation of $g(x)$ separately and multiply it with the differentiation of $f(g(x))$.
Students should keep in mind that a and b are constants and should not apply product rule within the bracket.
Also, it is recommended not to directly apply chain rule, first find differentiation of the terms within and then do substitution in the chain rule.