Question

Find \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]
\[y=\sin x\cos x\]

Answer 1

Hint:If u and v are two differentiable functions of x then \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\times \dfrac{du}{dx}-u\times \dfrac{dv}{dx}}{{{v}^{2}}}\]and this formula is called quotient rule. In this problem u and v are two differentiable functions of x so apply the quotient rule. Here they asked us to find the \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]so we have to apply the quotient rule two times then we will get the required answer.

Complete step-by-step answer:
Given that \[y=\sin x\cos x\]
We have to find \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]
Given \[y=\sin x\cos x\]
\[y=\dfrac{2\sin x\cos x}{2}\]
\[y=\dfrac{\sin 2x}{2}\]
Now multiply and divide 2 with numerator and denominator respectively then we will get,

We know that the formula of \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)\]is given by \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\times \dfrac{du}{dx}-u\times \dfrac{dv}{dx}}{{{v}^{2}}}\]
Now apply the above formula then we will get,
\[\dfrac{dy}{dx}=\dfrac{(2)\left( 2\cos 2x \right)-\left( \sin 2x \right)\left( 0 \right)}{{{(2)}^{2}}}\]. . . . . . . . . . . . (1)
\[\dfrac{dy}{dx}=\dfrac{4\cos 2x}{(4)}\]
\[\dfrac{dy}{dx}=\cos 2x\]. . . . . . . . . . . . . . . . (2)
Now again apply derivative to it then we will get the second derivative or double derivative.
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{\cos 2x}{1} \right)\]. . . . . . . . . . .. . . . . . . . . . . . . . . . . . (3)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{(1)\left( -2\sin 2x \right)-\left( \cos 2x \right)\left( 0 \right)}{{{(1)}^{2}}}\]. . . . . . . . . . . . . . . .(4)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-2\sin 2x\]. . . . . . . . . . . . (5)
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-2(2\sin x\cos x)\]
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-4\sin x\cos x\]

Note: The v can also be integer in quotient rule so we have applied quotient rule. The derivative of \[\sin ax\]is equal to \[a\cos ax\]and the derivative of \[\cos ax\]is equal to \[-a\sin ax\]where a is a positive integer. We know that the formulae of \[\sin 2x\]is equal to \[2\sin x\cos x\]. So we should be keen on basic trigonometric formulas and their derivative formulas for doing this problem. Carefully do the basic mathematical operations like addition subtraction then we will get the required answer.