Question
Question: If \(f\left( x \right) = 10\cos x + \left( {13 + 2x} \right)\sin x\), then \(f''\left( x \right) + f...
If f(x)=10cosx+(13+2x)sinx, then f′′(x)+f(x) is equal to:
A. cosx
B. 4cosx
C. sinx
D. 4sinx
Solution
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(x)=10cosx+(13+2x)sinx involves a product term, so we will have to apply product rule of differentiation in the process of differentiating f(x)=10cosx+(13+2x)sinx . Also derivatives of basic algebraic and trigonometric functions must be remembered thoroughly.
Complete step by step answer:
So, we have, f(x)=10cosx+(13+2x)sinx
Opening the brackets, we get,
⇒f(x)=10cosx+13sinx+2xsinx
Differentiating both sides of the equation, we get,
⇒f′(x)=dxd[10cosx+13sinx+2xsinx]
⇒f′(x)=dxd(10cosx)+dxd(13sinx)+dxd(2xsinx)
Taking constant out of the differentiation, we get,
⇒f′(x)=10dxd(cosx)+13dxd(sinx)+2dxd(xsinx)
Now, we know the derivative of cosx is −sinx and the derivative of sinx is cosx. So, we get,
⇒f′(x)=10(−sinx)+13(cosx)+2dxd(xsinx)
Using the product rule of differentiation dxd(f(x)×g(x))=f(x)×dxd(g(x))+g(x)×dxd(f(x)),
⇒f′(x)=10(−sinx)+13(cosx)+2[xdxd(sinx)+sinxdxd(x)]
Now, we know the derivative of x with respect to x is 1 and the derivative of sinx with respect to x is cosx. So, we get,
⇒f′(x)=10(−sinx)+13(cosx)+2[xcosx+sinx]
Opening brackets, we get,
⇒f′(x)=−10sinx+13cosx+2xcosx+2sinx
Simplifying the expression,
⇒f′(x)=13cosx+2xcosx−8sinx
Now, differentiate the above equation again with respect to x to find the second derivative of the function.
⇒f′′(x)=dxdy[13cosx+2xcosx−8sinx]
⇒f′′(x)=dxd(13cosx)+dxd(2xcosx)−dxd(8sinx)
Taking constant out of the differentiation, we get,
⇒f′′(x)=13dxd(cosx)+2dxd(xcosx)−8dxd(sinx)
Now, we know the derivative of cosx is −sinx and the derivative of sinx is cosx. So, we get,
⇒f′′(x)=13(−sinx)+2dxd(xcosx)−8cosx
Using the product rule of differentiation dxd(f(x)×g(x))=f(x)×dxd(g(x))+g(x)×dxd(f(x)),
⇒f′′(x)=−13sinx−8cosx+2[xdxd(cosx)+cosxdxd(x)]
Substituting the known derivatives again,
⇒f′′(x)=−13sinx−8cosx+2[x(−sinx)+cosx(1)]
Simplifying the expression, we get,
⇒f′′(x)=−13sinx−8cosx−2xsinx+2cosx
⇒f′′(x)=−13sinx−6cosx−2xsinx
Now, we substitute the value of function and its second derivative into the expression which has to be evaluated. So, we have,
f′′(x)+f(x)=−13sinx−6cosx−2xsinx+10cosx+13sinx+2xsinx
Adding up the like terms and cancelling similar terms with opposite signs, we get,
⇒f′′(x)+f(x)=−6cosx+10cosx
∴f′′(x)+f(x)=4cosx
So, the value of the expression f′′(x)+f(x) for f(x)=10cosx+(13+2x)sinx is 4cosx.
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.