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Question: In \(x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4(y>0)\) , then \(\dfrac{dy}{dx}at\text{ }x=e\) is equal to: ...

In xln⁡(ln⁡x)−x2+y2=4(y>0)x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4(y>0) , then dydxat x=e\dfrac{dy}{dx}at\text{ }x=e is equal to:
A. e4+e2\dfrac{e}{\sqrt{4+{{e}^{2}}}}
B. 1+2e24+e2\dfrac{1+2e}{2\sqrt{4+{{e}^{2}}}}
C. 2e−124+e2\dfrac{2e-1}{2\sqrt{4+{{e}^{2}}}}
D. 1+2e4+e2\dfrac{1+2e}{\sqrt{4+{{e}^{2}}}}

Explanation

Solution

The required expression can be obtained by simply differentiating the given equation; xln⁡(ln⁡x)−x2+y2=4x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4 with respect to x implicitly i.e. dependent variable appears in the derivative and wise substitution of values at different steps of solution.

Complete step by step answer:
Given equation xln⁡(ln⁡x)−x2+y2=4x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4 can be differentiated with respect to x implicitly.
xln⁡(ln⁡x)−x2+y2=4…eq(1)x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4\ldots eq(1)
On rearranging eq(1)eq\left( 1 \right)we get,
y2=x2−xln⁡(ln⁡x)+4…eq(2){{y}^{2}}={{x}^{2}}-x\ln (\ln x)+4\ldots eq(2)
We know differentiate on both side with respect to x. Differentiating above equation need following important formula:
d(xn)dx=nxn−1\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}, d(f(x).g(x))dx=f′(x)g(x)+f(x)g′(x)\dfrac{d\left( f\left( x \right).g\left( x \right) \right)}{dx}=f'\left( x \right)g\left( x \right)+f\left( x \right)g'\left( x \right), d(f(g(x))dx=f′(g(x))g′(x)\dfrac{d(f\left( g\left( x \right) \right)}{dx}={f}'\left( g\left( x \right) \right){g}'\left( x \right), d(xln⁡(ln⁡(x)))dx=ln⁡(ln⁡(x))+1ln⁡(x)\dfrac{d\left( x\ln \left( \ln \left( x \right) \right) \right)}{dx}=\ln \left( \ln \left( x \right) \right)+\dfrac{1}{\ln \left( x \right)}, d(c)dx=0\dfrac{d\left( c \right)}{dx}=0
ddx(y2)=ddx(x2−xln⁡(ln⁡(x))+4)\dfrac{d}{dx}({{y}^{2}})=\dfrac{d}{dx}({{x}^{2}}-x\ln (\ln (x))+4)
ddx(y2)=ddx(x2)−ddx[xln⁡(ln⁡(x))]+ddx(4)\dfrac{d}{dx}\left( {{y}^{2}} \right)=\dfrac{d}{dx}\left( {{x}^{2}} \right)-\dfrac{d}{dx}\left[ x\ln (\ln (x)) \right]+\dfrac{d}{dx}\left( 4 \right)
2ydydx=2x−ln⁡(ln⁡x)−1ln⁡x+02y\dfrac{dy}{dx}=2x-\ln (\ln x)-\dfrac{1}{\ln x}+0
dydx=12y(2x−ln⁡(ln⁡x)−1ln⁡x)…eq(3)\dfrac{dy}{dx}=\dfrac{1}{2y}\left( 2x-\ln (\ln x)-\dfrac{1}{\ln x} \right)\ldots eq(3)
We can obtain the value of y from the original equation xln⁡(ln⁡x)−x2+y2=4x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4by substituting the value of x=ex=e,
eln(ln(e))−e2+y2=4eln(ln(e))-{{e}^{2}}+{{y}^{2}}=4
As we know thatln⁡(ln⁡(e))=1,ln⁡(1)=0\ln (\ln (e))=1,\ln (1)=0 we get following value of y,
y=e2+4y=\sqrt{{{e}^{2}}+4}
Atx=ex=e, the eq(3)eq\left( 3 \right)gives,
dydx=12y(2e−1)\dfrac{dy}{dx}=\dfrac{1}{2y}\left( 2e-1 \right)
dydx=2e−12e2+4at x=e\dfrac{dy}{dx}=\dfrac{2e-1}{2\sqrt{{{e}^{2}}+4}}at\text{ }x=e

So, the correct answer is “Option C”.

Note: The implicit differentiation must be done properly. The equation xln⁡(ln⁡x)−x2+y2=4x\ln (\ln x)-{{x}^{2}}+{{y}^{2}}=4can be implicitly differentiated without rearranging the terms (rearranging is just for simplicity) as given below:
ddx(xln⁡(ln⁡x)−x2+y2)=ddx(4)\dfrac{d}{dx}(x\ln (\ln x)-{{x}^{2}}+{{y}^{2}})=\dfrac{d}{dx}(4)
1ln⁡x+ln⁡(ln⁡(x))−2x+2ydydx=0\dfrac{1}{\ln x}+\ln (\ln (x))-2x+2y\dfrac{dy}{dx}=0