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Let $y^3 - y + x = 0$ then $\frac{\mathrm{d} y}{\mathrm{d} x}$ = ?

Option 1)
$1 - 3y^{2}$

Option 2)
$\frac{1}{1 - 3y^{2}}$

Option 3)
$\frac{1}{3y^{2}-1}$

Option 4)
$3y^{2} -1$

Answers (1)

best_answer

As we have learnt,

Derivative of implict function -

When y is given in any function then we find derivative of function first then find derivative of y and collect the terms containing

dy / dx on left side and find dy / dx in terms of x & y

- wherein

$ex:$

$Let \:\:y=siny$

$\frac{dy}{dx}=cosy\times\frac{dy}{dx}$

$\frac{dy}{dx}(1-cosy)=0$

$ex:$

$Let\:\:y=sin(xy)$

$\frac{dy}{dx}=cos(xy)\times(1.y+\frac{xdy}{dx})$

$\frac{dy}{dx}(1-x\:cos(xy))=y\:cos(xy)$

$\therefore \:\frac{dy}{dx}=\frac{y\:cos(xy)}{1-x\:cos(xy)}$

$\frac{\mathrm{d} y^3 - y + x}{\mathrm{d} x} = 0 \Rightarrow 3y^2 \frac{\mathrm{d} y}{\mathrm{d} x} - \frac{\mathrm{d} y}{\mathrm{d} x} + 1 = 0$

$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}(3y^2 - 1) = -1 \Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{1}{1 - 3y^2}$

Option 1)

$1 - 3y^{2}$

Option 2)

$\frac{1}{1 - 3y^{2}}$

Option 3)

$\frac{1}{3y^{2}-1}$

Option 4)

$3y^{2} -1$

Posted by

Himanshu

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