Question

Find the value of the derivative $\dfrac{d}{dx}\left( \cos \left( a{{x}^{2}}+bx+c \right) \right)$.
A. $-\left( 2ax+b \right)\sin \left( a{{x}^{2}}+bx+c \right)$
B. $-\left( ax+b \right)\sin \left( a{{x}^{2}}-bx-c \right)$
C. $\left( 2ax+b \right)\sin \left( a{{x}^{2}}+bx+c \right)$
D. $-\left( ax+b \right)\sin \left( a{{x}^{2}}+bx+c \right)$

Answer 1

Hint: To solve this question, we should know the chain rule of differentiation. Let us consider a function $y=f\left( g\left( x \right) \right)$ and the derivative of y with respect to x is given by the formula $\dfrac{dy}{dx}=f'\left( g\left( x \right) \right)\times g'\left( x \right)$. In our question, we have the function f as the cosine function and the function g as the quadratic expression $a{{x}^{2}}+bx+c$. Using the above formula, we can find the required derivative.

Complete step-by-step solution:
Let us consider a function of x such that $y=f\left( g\left( x \right) \right)$ and let us consider the derivative of this function. We will consider $g\left( x \right)=u$. The function of y becomes
$y=f\left( u \right)$
We can write the derivative of the above term with respect to x as
$\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$
We can substitute the assumed terms in the above derivative to get
$\dfrac{dy}{dx}=\dfrac{d\left( f\left( u \right) \right)}{du}.\dfrac{d\left( g\left( x \right) \right)}{dx}$
Now, we can write the above differential as
$\dfrac{dy}{dx}=f'\left( u \right).g'\left( x \right)$
Substituting the term $g\left( x \right)=u$ back in the derivative, we get
\[\dfrac{dy}{dx}=f'\left( g\left( x \right) \right).g'\left( x \right)\]
This is the rule known as the chain rule in differentiation. Now, let us consider the derivative given to us in the question.
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \cos \left( a{{x}^{2}}+bx+c \right) \right)$
Comparing the above derivative with the derivative that we got in the chain rule, we get
$y=f\left( g\left( x \right) \right)=\cos \left( a{{x}^{2}}+bx+c \right)$
$\begin{align}
  & f\left( u \right)=\cos u \\
 & u=g\left( x \right)=a{{x}^{2}}+bx+c \\
\end{align}$
Using the chain rule, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( \cos \left( a{{x}^{2}}+bx+c \right) \right)=\dfrac{d\left( \cos u \right)}{du}.\dfrac{d\left( a{{x}^{2}}+bx+c \right)}{dx}$
We know the differentiation formula
$\begin{align}
  & \dfrac{d\left( \cos y \right)}{dy}=-\sin u \\
 & \dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} \\
\end{align}$
Using these two relations, we get
$\dfrac{d\left( \cos u \right)}{du}.\dfrac{d\left( a{{x}^{2}}+bx+c \right)}{dx}=-\sin u.\left( 2ax+b \right)$
Substituting $u=g\left( x \right)=a{{x}^{2}}+bx+c$ in the above equation, we get
$\dfrac{d}{dx}\left( \cos \left( a{{x}^{2}}+bx+c \right) \right)=-\sin \left( a{{x}^{2}}+bx+c \right).\left( 2ax+b \right)$
$\therefore $The value of the given differentiation is$\dfrac{d}{dx}\left( \cos \left( a{{x}^{2}}+bx+c \right) \right)=-\sin \left( a{{x}^{2}}+bx+c \right).\left( 2ax+b \right)$. The answer is option-A

Note: Some students might forget the negative sign in the derivative of cosine term which leads to a wrong answer. We can directly apply the chain rule on this question by ignoring what is inside the cosine term and directly differentiating it as if we differentiate $\cos x$ and after that, we should multiply this derivative by the derivative of the term inside the cosine function. The chain rule is a very important rule in differentiation which should be remembered.