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Let y^3 -yx + 3 = 0 then slope of tangent at (10, 3) will be?

Option: 1

\frac{3}{14}


Option: 2

\frac{3}{16}


Option: 3

\frac{3}{17}


Option: 4

\frac{3}{19}


Answers (1)

best_answer

As we have learnt,

 

Slope of the tangent -

Let y = f(x) is a curve then  dy / dx = f'(x) and at a particular point (h, k) it gives slope of tangent. From fig

M_{T}=\lim_{\delta x\rightarrow 0}\:\frac{(y+\delta y)-y}{(x+\delta x)-x}=\lim_{\delta x\rightarrow 0}\:\frac{\delta y}{\delta x}

- wherein

 

 On differenciating both sides, we have,

3y^2\frac{\mathrm{d} y}{\mathrm{d} x} -\left(y+x\frac{\mathrm{d} y}{\mathrm{d} x} \right ) +0= 0 \Rightarrow (3y^2 -x )\frac{\mathrm{d} y}{\mathrm{d} x} = y \\*\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{y}{3y^2 - x}

\therefore\frac{\mathrm{d} }{\mathrm{d} x}\;at\;(10,3) = \frac{3}{27-10} =\frac{3}{17}

 

Posted by

Rishi

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