Question

let 5f(x)+ 3f(1/x)=x+2 and y= xf(x) . then x=1 point dy/dx=?​

Answer 1

GIVEN :

Let $5f(x)+ 3f(\frac{1}{x})=x+2$ and y= xf(x) . then x=1 point $\frac{dy}{dx}$

TO FIND :

The value of $\frac{dy}{dx}$ at x=1

SOLUTION :

Given that $5f(x)+ 3f(\frac{1}{x})=x+2$ and y= xf(x)

Let $5f(x)+ 3f(\frac{1}{x})=x+2\hfill (1)$

Now replace $x=\frac{1}{x}$

$5f(\frac{1}{x})+ 3f(x)=\frac{1}{x}+2\hfill (2)$

Multiply the equation (1) into 5 we get,

$25f(x)+ 15f(\frac{1}{x})=5x+10\hfill (3)$

Multiply the equation (2) into 3 we get,

$15f(\frac{1}{x})+9f(x)=3(\frac{1}{x})+6\hfill (4)$

Subtracting the equations (3) and (4) we get

$25f(x)+ 15f(\frac{1}{x})=5x+10$

$15f(\frac{1}{x})+9f(x)=3(\frac{1}{x})+6$   (-)

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$16f(x)=5x-3(\frac{1}{x})+4$

$16f(x)=\frac{5x^2-3+4x}{x}$

$xf(x)=\frac{5x^2+4x-3}{16}$

$y=xf(x)=\frac{5x^2+4x-3}{16}$

$y=\frac{5x^2+4x-3}{16}$

Differentiating y with respect to x,

$\frac{dy}{dx}=\frac{2(5x)^{2-1}+4(1)-0}{16}$

By using the differentiation formulae :

i) $\frac{d(x^n)}{dx}=nx^{n-1}$

ii) $\frac{d(ax)}{dx}=a$

iii) $\frac{d(c)}{dx}=0$

$\frac{dy}{dx}=\frac{10x+4}{16}$

Now at x=1 for $\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{10(1)+4}{16}$

$=\frac{14}{16}$

$=\frac{7}{8}$

∴ the value of $\frac{dy}{dx}$ at x=1 is $\frac{7}{8}$

Answer 2

SOLUTION

GIVEN

$\displaystyle \sf{ 5f(x) + 3f \bigg( \frac{1}{x} \bigg) = x + 2\: \: \: and \: \: y = xf(x) }$

TO DETERMINE

$\displaystyle \sf{ \frac{dy}{dx} \: \: \: \: \: at \: \: x = 1}$

EVALUATION

Here it is given that

$\displaystyle \sf{ 5f(x) + 3f \bigg( \frac{1}{x} \bigg) = x + 2\: \: \: ....(1)}$

Putting x = 1 in both sides we get

$\displaystyle \sf{ 5f(1) + 3f \bigg( \frac{1}{1} \bigg) = 1 + 2\: }$

$\implies \displaystyle \sf{ 5f(1) + 3f (1) = 3\: }$

$\implies \displaystyle \sf{ 8f(1) = 3\: }$

$\implies \displaystyle \sf{ f(1) = \frac{3}{8} \: }$

Differentiating both sides of Equation (1) we get

$\displaystyle \sf{ 5f'(x) - \frac{3}{ {x}^{2} } \: f' \bigg( \frac{1}{x} \bigg) = 1\: \: \: }$

$\implies \displaystyle \sf{ 5f'(1) - \frac{3}{ {(1)}^{2} } \: f' \bigg( \frac{1}{1} \bigg) = 1\: \: \: }$

$\implies \displaystyle \sf{ 2f'(1) = 1\: \: \: }$

$\implies \displaystyle \sf{ f'(1) = \frac{1}{2} \: \: \: }$

Here

$\sf{ y = xf(x)\: }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ \frac{dy}{dx} = f(x) + xf'(x)}$

Putting x = 1 in both sides we get

$\displaystyle \sf{ \frac{dy}{dx} \bigg|_{x = 1} = f(1) + 1.f'(1)}$

$\implies \displaystyle \sf{ \frac{dy}{dx} \bigg|_{x = 1} = f(1) +f'(1)}$

$\implies \displaystyle \sf{ \frac{dy}{dx} \bigg|_{x = 1} = \frac{3}{8} + \frac{1}{2} }$

$\implies \displaystyle \sf{ \frac{dy}{dx} \bigg|_{x = 1} = \frac{7}{8} }$

FINAL ANSWER

$\displaystyle \sf{ \frac{dy}{dx} \bigg|_{x = 1} = \frac{7}{8} }$

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