If (a+bx)^{e^{y/x}} = x, then d²y/dx² =

Correct Answer is : 1x3(xy′−y)21/x³ (x y' - y)²x31(xy′−y)2

Correct Answer is : 1x3(xy′−y)21/x³ (x y' - y)²x31(xy′−y)2

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TS EAMCET 4-Aug-2021 Shift 2 Question Paper

Section: Mathematics

Question : 69 of 160

Marks: +1, -0

(a+bx)^{e^{\frac{y}{x}}}

x

\frac{d^2y}{dx^2}

Solution:

(a+b x) e^{\frac{y}{x}} = x

Taking log on both the sides, we get

\log (a+b x) + \; \frac{y}{x} = \log x

y = x [\log x - \log (a+b x)]

.....(i)

Differentiating w.r.t.

x

y' = x \left( \; \frac{1}{x} - \; \frac{b}{a+b x} \right) + [\log x - \log (a+b x)]

y' = 1 - \; \frac{xb}{a+b x} - \; \frac{y}{x}

...(ii)

\; \frac{b}{a+b x} = \; \frac{1}{x} + \; \frac{y}{x^2} - \; \frac{y'}{x}

....(iii)

Again, differentiating Eq. (ii) w.r.t.

x

y'' = -b \left[ \; \frac{(a+b x) - b x}{(a+b x)^2} \right] + \; \frac{y'}{x} - \; \frac{y}{x^2}

\;\; = \; \frac{x b^2}{(a+b x)^2} - \; \frac{b}{a+b x} + \; \frac{y'}{x} - \; \frac{y}{x^2}

From Eq. (iii),

\; x \left( \; \frac{1}{x} + \; \frac{y}{x^2} - \; \frac{y'}{x} \right)^2 - \left( \; \frac{1}{x} + \; \frac{y}{x} - \; \frac{y'}{x} \right) + \; \frac{y'}{x} - \; \frac{y}{x^2}

= \;\; \; \frac{y^2}{x^3} + \; \frac{y'^2}{x} - \; \frac{2 y y'}{x^2}

= \;\; \; \frac{1}{x^3} \left[ y^2 + (x y)^2 - 2 x y y' \right]

= \;\; \; \frac{1}{x^3} (x y' - y)^2

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