Question

if x3+5x2y+xy-5=0 then dy/dx is​

Answer 1

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: {x}^{3} + {5x}^{2}y + xy - 5 = 0$

can be rewritten as

$\rm :\longmapsto\: {x}^{3} + ({5x}^{2} + x)y - 5 = 0$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} \bigg[ {x}^{3} + ({5x}^{2} + x)y - 5 \bigg]= 0$

$\rm :\longmapsto\:\dfrac{d}{dx} {x}^{3} + \dfrac{d}{dx} (5 {x}^{2} + x)y - \dfrac{d}{dx} 5 = 0$

We know,

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} }}$

$\boxed{\tt{ \dfrac{d}{dx} k = 0}}$

$\boxed{\tt{ \dfrac{d}{dx} uv = u\dfrac{d}{dx} v + v\dfrac{d}{dx} u}}$

So, using these Identities, we get

$\rm :\longmapsto\: {3x}^{3 - 1} + ( {5x}^{2} + x)\dfrac{d}{dx} y + y\dfrac{d}{dx} ( {5x}^{2} + x) - 0 = 0$

$\rm :\longmapsto\: {3x}^{2} + ( {5x}^{2} + x)\dfrac{dy}{dx} + y( {10x}^{2 - 1} + 1) = 0$

$\rm :\longmapsto\: {3x}^{2} + ( {5x}^{2} + x)\dfrac{dy}{dx} + y( 10x + 1) = 0$

$\rm :\longmapsto\:( {5x}^{2} + x)\dfrac{dy}{dx} = - {3x}^{2} - y(10x + 1)$

$\bf\implies \:\dfrac{dy}{dx} = - \dfrac{ {3x}^{2} + y(10x + 1)}{ {5x}^{2} + x}$

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MORE TO LEARN

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given function is

$\rm :\longmapsto\: {x}^{3} + {5x}^{2}y + xy - 5 = 0$

can be rewritten as

$\rm :\longmapsto\: {x}^{3} + ({5x}^{2} + x)y - 5 = 0$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} \bigg[ {x}^{3} + ({5x}^{2} + x)y - 5 \bigg]= 0$

$\rm :\longmapsto\:\dfrac{d}{dx} {x}^{3} + \dfrac{d}{dx} (5 {x}^{2} + x)y - \dfrac{d}{dx} 5 = 0$

We know,

$\boxed{\tt{ \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} }}$

$\boxed{\tt{ \dfrac{d}{dx} k = 0}}$

$\boxed{\tt{ \dfrac{d}{dx} uv = u\dfrac{d}{dx} v + v\dfrac{d}{dx} u}}$

So, using these Identities, we get

$\rm :\longmapsto\: {3x}^{3 - 1} + ( {5x}^{2} + x)\dfrac{d}{dx} y + y\dfrac{d}{dx} ( {5x}^{2} + x) - 0 = 0$

$\rm :\longmapsto\: {3x}^{2} + ( {5x}^{2} + x)\dfrac{dy}{dx} + y( {10x}^{2 - 1} + 1) = 0$

$\rm :\longmapsto\: {3x}^{2} + ( {5x}^{2} + x)\dfrac{dy}{dx} + y( 10x + 1) = 0$

$\rm :\longmapsto\:( {5x}^{2} + x)\dfrac{dy}{dx} = - {3x}^{2} - y(10x + 1)$

$\bf\implies \:\dfrac{dy}{dx} = - \dfrac{ {3x}^{2} + y(10x + 1)}{ {5x}^{2} + x}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

MORE TO LEARN

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$