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Question: If \(y = {x^3}\log x\) , prove that \(\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{6}{x}\)....

If y=x3log⁡xy = {x^3}\log x , prove that d4ydx4=6x\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{6}{x}.

Explanation

Solution

It is given in the question that y=x3log⁡xy = {x^3}\log x .
Then, we have to prove d4ydx4=6x\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{6}{x} .
First, to find dydx\dfrac{{dy}}{{dx}} differentiate y=x3log⁡xy = {x^3}\log x with respect to x. Then, differentiate dydx\dfrac{{dy}}{{dx}} to get d2ydx2\dfrac{{{d^2}y}}{{d{x^2}}} and repeat the same steps two times to get d4ydx4\dfrac{{{d^4}y}}{{d{x^4}}}.

Complete step by step solution:
It is given in the question that y=x3log⁡xy = {x^3}\log x .
Then, we have to prove d4ydx4=6x\dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{6}{x} .
First, we have to differentiate y=x3log⁡xy = {x^3}\log x with respect to x.
⇒dydx=ddx(x3log⁡x)\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {{x^3}\log x} \right)
⇒dydx=x3ddxlog⁡x+log⁡xddxx3\Rightarrow \dfrac{{dy}}{{dx}} = {x^3}\dfrac{d}{{dx}}\log x + \log x\dfrac{d}{{dx}}{x^3}
⇒dydx=x3x+3x2log⁡x\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{x^3}}}{x} + 3{x^2}\log x
⇒dydx=x2(1+3log⁡x)\Rightarrow \dfrac{{dy}}{{dx}} = {x^2}\left( {1 + 3\log x} \right)
Now, to find the second derivative we have to differentiate dydx=x2(1+3log⁡x)\dfrac{{dy}}{{dx}} = {x^2}\left( {1 + 3\log x} \right) with respect to x.
⇒d2ydx2=x2ddx(1+3log⁡x)+(1+3log⁡x)ddxx2\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = {x^2}\dfrac{d}{{dx}}\left( {1 + 3\log x} \right) + \left( {1 + 3\log x} \right)\dfrac{d}{{dx}}{x^2}
⇒d2ydx2=x2×3x+(1+3log⁡x)2x\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = {x^2} \times \dfrac{3}{x} + \left( {1 + 3\log x} \right)2x
⇒d2ydx2=x2×3x+(1+3log⁡x)2x\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = {x^2} \times \dfrac{3}{x} + \left( {1 + 3\log x} \right)2x
⇒d2ydx2=x+(5+6log⁡x)\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = x + \left( {5 + 6\log x} \right)
Now, to find the third derivative we have to differentiate d2ydx2=x+(5+6log⁡x)\dfrac{{{d^2}y}}{{d{x^2}}} = x + \left( {5 + 6\log x} \right) with respect to x.
⇒d3ydx3=xddx(5+6log⁡x)+(5+6log⁡x)ddxx\Rightarrow \dfrac{{{d^3}y}}{{d{x^3}}} = x\dfrac{d}{{dx}}\left( {5 + 6\log x} \right) + \left( {5 + 6\log x} \right)\dfrac{d}{{dx}}x
⇒d3ydx3=x×6x+(5+6log⁡x)\Rightarrow \dfrac{{{d^3}y}}{{d{x^3}}} = x \times \dfrac{6}{x} + \left( {5 + 6\log x} \right)
⇒d3ydx3=11+6log⁡x\Rightarrow \dfrac{{{d^3}y}}{{d{x^3}}} = 11 + 6\log x
Now, to find the fourth derivative we have to differentiate d3ydx3=11+6log⁡x\dfrac{{{d^3}y}}{{d{x^3}}} = 11 + 6\log x with respect to x.
⇒d4ydx4=6x\Rightarrow \dfrac{{{d^4}y}}{{d{x^4}}} = \dfrac{6}{x}
Hence proved.

Note:
Product Rule: The product rule is a formula used to find the derivatives of products of two or more functions. It may be stated as
(f.g)′=f′.g+f.g′\left( {f.g} \right)' = f'.g + f.g'
ddx(u.v)=dudx.v+u.dvdx\dfrac{d}{{dx}}\left( {u.v} \right) = \dfrac{{du}}{{dx}}.v + u.\dfrac{{dv}}{{dx}}