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

Find d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}
y=sin⁡xcos⁡xy=\sin x\cos x

Explanation

Solution

Hint:If u and v are two differentiable functions of x then ddx(uv)=v×dudx−u×dvdxv2\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 d2ydx2\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⁡xcos⁡xy=\sin x\cos x
We have to find d2ydx2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}
Given y=sin⁡xcos⁡xy=\sin x\cos x
y=2sin⁡xcos⁡x2y=\dfrac{2\sin x\cos x}{2}
y=sin⁡2x2y=\dfrac{\sin 2x}{2}
Now multiply and divide 2 with numerator and denominator respectively then we will get,

We know that the formula of ddx(uv)\dfrac{d}{dx}\left( \dfrac{u}{v} \right)is given by ddx(uv)=v×dudx−u×dvdxv2\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,
dydx=(2)(2cos⁡2x)−(sin⁡2x)(0)(2)2\dfrac{dy}{dx}=\dfrac{(2)\left( 2\cos 2x \right)-\left( \sin 2x \right)\left( 0 \right)}{{{(2)}^{2}}}. . . . . . . . . . . . (1)
dydx=4cos⁡2x(4)\dfrac{dy}{dx}=\dfrac{4\cos 2x}{(4)}
dydx=cos⁡2x\dfrac{dy}{dx}=\cos 2x. . . . . . . . . . . . . . . . (2)
Now again apply derivative to it then we will get the second derivative or double derivative.
d2ydx2=ddx(cos⁡2x1)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( \dfrac{\cos 2x}{1} \right). . . . . . . . . . .. . . . . . . . . . . . . . . . . . (3)
d2ydx2=(1)(−2sin⁡2x)−(cos⁡2x)(0)(1)2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{(1)\left( -2\sin 2x \right)-\left( \cos 2x \right)\left( 0 \right)}{{{(1)}^{2}}}. . . . . . . . . . . . . . . .(4)
d2ydx2=−2sin⁡2x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-2\sin 2x. . . . . . . . . . . . (5)
d2ydx2=−2(2sin⁡xcos⁡x)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-2(2\sin x\cos x)
d2ydx2=−4sin⁡xcos⁡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\sin axis equal to acos⁡axa\cos axand the derivative of cos⁡ax\cos axis equal to −asin⁡ax-a\sin axwhere a is a positive integer. We know that the formulae of sin⁡2x\sin 2xis equal to 2sin⁡xcos⁡x2\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.