Question

If $f\left( x \right) = 10\cos x + \left( {13 + 2x} \right)\sin x$, then $f''\left( x \right) + f\left( x \right)$ is equal to:
A. $\cos x$
B. $4\cos x$
C. $\sin x$
D. $4\sin x$

Answer 1

Hint:In the given problem, we are required to find the value of the expressions involving the second order derivative of the given function. Since, $f\left( x \right) = 10\cos x + \left( {13 + 2x} \right)\sin x$ involves a product term, so we will have to apply product rule of differentiation in the process of differentiating $f\left( x \right) = 10\cos x + \left( {13 + 2x} \right)\sin x$ . Also derivatives of basic algebraic and trigonometric functions must be remembered thoroughly.

Complete step by step answer:
So, we have, $f\left( x \right) = 10\cos x + \left( {13 + 2x} \right)\sin x$
Opening the brackets, we get,
$ \Rightarrow f\left( x \right) = 10\cos x + 13\sin x + 2x\sin x$
Differentiating both sides of the equation, we get,
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left[ {10\cos x + 13\sin x + 2x\sin x} \right]$
\[ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {10\cos x} \right) + \dfrac{d}{{dx}}\left( {13\sin x} \right) + \dfrac{d}{{dx}}\left( {2x\sin x} \right)\]

Taking constant out of the differentiation, we get,
\[ \Rightarrow f'\left( x \right) = 10\dfrac{d}{{dx}}\left( {\cos x} \right) + 13\dfrac{d}{{dx}}\left( {\sin x} \right) + 2\dfrac{d}{{dx}}\left( {x\sin x} \right)\]
Now, we know the derivative of $\cos x$ is $ - \sin x$ and the derivative of $\sin x$ is $\cos x$. So, we get,
\[ \Rightarrow f'\left( x \right) = 10\left( { - \sin x} \right) + 13\left( {\cos x} \right) + 2\dfrac{d}{{dx}}\left( {x\sin x} \right)\]
Using the product rule of differentiation $\dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right) + g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right)$,
\[ \Rightarrow f'\left( x \right) = 10\left( { - \sin x} \right) + 13\left( {\cos x} \right) + 2\left[ {x\dfrac{d}{{dx}}\left( {\sin x} \right) + \sin x\dfrac{{d\left( x \right)}}{{dx}}} \right]\]

Now, we know the derivative of $x$ with respect to x is $1$ and the derivative of $\sin x$ with respect to x is $\cos x$. So, we get,
\[ \Rightarrow f'\left( x \right) = 10\left( { - \sin x} \right) + 13\left( {\cos x} \right) + 2\left[ {x\cos x + \sin x} \right]\]
Opening brackets, we get,
\[ \Rightarrow f'\left( x \right) = - 10\sin x + 13\cos x + 2x\cos x + 2\sin x\]
Simplifying the expression,
\[ \Rightarrow f'\left( x \right) = 13\cos x + 2x\cos x - 8\sin x\]
Now, differentiate the above equation again with respect to x to find the second derivative of the function.
\[ \Rightarrow f''\left( x \right) = \dfrac{{dy}}{{dx}}\left[ {13\cos x + 2x\cos x - 8\sin x} \right]\]
\[ \Rightarrow f''\left( x \right) = \dfrac{d}{{dx}}\left( {13\cos x} \right) + \dfrac{d}{{dx}}\left( {2x\cos x} \right) - \dfrac{d}{{dx}}\left( {8\sin x} \right)\]
Taking constant out of the differentiation, we get,
\[ \Rightarrow f''\left( x \right) = 13\dfrac{d}{{dx}}\left( {\cos x} \right) + 2\dfrac{d}{{dx}}\left( {x\cos x} \right) - 8\dfrac{d}{{dx}}\left( {\sin x} \right)\]

Now, we know the derivative of $\cos x$ is $ - \sin x$ and the derivative of $\sin x$ is $\cos x$. So, we get,
\[ \Rightarrow f''\left( x \right) = 13\left( { - \sin x} \right) + 2\dfrac{d}{{dx}}\left( {x\cos x} \right) - 8\cos x\]
Using the product rule of differentiation $\dfrac{d}{{dx}}\left( {f(x) \times g(x)} \right) = f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right) + g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right)$,
\[ \Rightarrow f''\left( x \right) = - 13\sin x - 8\cos x + 2\left[ {x\dfrac{d}{{dx}}\left( {\cos x} \right) + \cos x\dfrac{{d\left( x \right)}}{{dx}}} \right]\]
Substituting the known derivatives again,
\[ \Rightarrow f''\left( x \right) = - 13\sin x - 8\cos x + 2\left[ {x\left( { - \sin x} \right) + \cos x\left( 1 \right)} \right]\]

Simplifying the expression, we get,
\[ \Rightarrow f''\left( x \right) = - 13\sin x - 8\cos x - 2x\sin x + 2\cos x\]
\[ \Rightarrow f''\left( x \right) = - 13\sin x - 6\cos x - 2x\sin x\]
Now, we substitute the value of function and its second derivative into the expression which has to be evaluated. So, we have,
$f''\left( x \right) + f\left( x \right) = - 13\sin x - 6\cos x - 2x\sin x + 10\cos x + 13\sin x + 2x\sin x$
Adding up the like terms and cancelling similar terms with opposite signs, we get,
$ \Rightarrow f''\left( x \right) + f\left( x \right) = - 6\cos x + 10\cos x$
$ \therefore f''\left( x \right) + f\left( x \right) = 4\cos x$
So, the value of the expression $f''\left( x \right) + f\left( x \right)$ for $f\left( x \right) = 10\cos x + \left( {13 + 2x} \right)\sin x$ is $4\cos x$.

Hence, the correct answer is option B.

Note: One must know the product rule of differentiation in order to solve the problem. The derivatives of basic trigonometric and algebraic functions must be learned by heart in order to find derivatives of complex composite functions using product rule and chain rule of differentiation. We should also know the simplification rule to simplify the expressions and find the value of the entity.