Implicit Differentiation

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GRE Quantitative Reasoning › Implicit Differentiation

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Differentiate the following with respect to .

$\cos x +xy^2-\sin y=y^2$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\sin x-y^2}{(2xy-\cos y -2y) }$

CORRECT

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{-(\sin x+y^2)}{2xy-\cos y -2y }$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\sin x-y^2}{xy-\cos y -y }$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\cos x-y^2}{(2xy-\sin y -2y) }$

Explanation

Our first step is to differentiate both sides with respect to :

$\cos x \frac{\mathrm{d} }{\mathrm{d} x} +xy^2\ \frac{\mathrm{d} }{\mathrm{d} x}-\sin y \ \frac{\mathrm{d} }{\mathrm{d} x}=y^2 \ \frac{\mathrm{d} }{\mathrm{d} x}$

Note: we can differentiate the terms that are functions of with respect to , just remember to multiply it by $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

$-\sin x +y^2 +2xy \ \frac{\mathrm{d} y}{\mathrm{d} x}-\cos y \ \frac{\mathrm{d} y}{\mathrm{d} x}=2y \ \frac{\mathrm{d}y }{\mathrm{d} x}$

Note: The product rule was applied above:

$xy^2 \ \frac{\mathrm{d} }{\mathrm{d} x}=x\frac{\mathrm{d} }{\mathrm{d} x}y^2+y^2\frac{\mathrm{d} y}{\mathrm{d} x}x=y^2+2yx\frac{\mathrm{d} y}{\mathrm{d} x}$

$2xy \ \frac{\mathrm{d} y}{\mathrm{d} x}-\cos y \ \frac{\mathrm{d} y}{\mathrm{d} x}-2y \ \frac{\mathrm{d}y }{\mathrm{d} x} =\sin x-y^2$

$(2xy-\cos y -2y) \ \frac{\mathrm{d}y }{\mathrm{d} x} =\sin x-y^2$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\sin x-y^2}{(2xy-\cos y -2y) }$

Find

$\frac{\mathrm{d} y}{\mathrm{d} x}$ for .

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-x}{3y^2}$

CORRECT

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{x}{3y^2}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-x}{3y}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{x^2}{2y^2}$

Explanation

Our first step would be to differentiate both sides with respect to :

$x^2 \frac{\mathrm{d} }{\mathrm{d} x}+2y^3 \frac{\mathrm{d} }{\mathrm{d} x}=8 \frac{\mathrm{d} }{\mathrm{d} x}$

The functions of can be differentiated with respect to , just remember to multiply by $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

$2x + 6y^2 \frac{\mathrm{d}y }{\mathrm{d} x}=0$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-2x}{6y^2}$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-x}{3y^2}$

Differentiate the following to solve for $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

$\cos y + 3x^3+4y=\sin x$

$\frac{\mathrm{d}y }{\mathrm{d} x}= \frac{(\cos x-9x^2)}{(4-\sin y )}$

CORRECT

$\frac{\mathrm{d}y }{\mathrm{d} x}= \frac{\cos x-9x^2}{(\sin y +4 )}$

$\frac{\mathrm{d}y }{\mathrm{d} x}= \frac{-\cos x+9x^2}{(\sin y +4y )}$

$\frac{\mathrm{d}y }{\mathrm{d} x}= \frac{\cos x-9x^2}{(\sin y -4 )}$

Explanation

Our first step is to differentiate both sides with respect to :

$\cos y \ \frac{\mathrm{d} }{\mathrm{d} x}+ 3x^3 \frac{\mathrm{d} }{\mathrm{d} x}+4y \frac{\mathrm{d} }{\mathrm{d} x}=\sin x \frac{\mathrm{d} }{\mathrm{d} x}$

The functions of can by differentiated with respect to , just remember to multiply them by $\frac{\mathrm{d} y}{\mathrm{d} x}$ :

$-\sin y \ \frac{\mathrm{d}y }{\mathrm{d} x}+ 9x^2 +4 \frac{\mathrm{d}y }{\mathrm{d} x}=\cos x$

$-\sin y \ \frac{\mathrm{d}y }{\mathrm{d} x}+4 \frac{\mathrm{d}y }{\mathrm{d} x}=\cos x-9x^2$

$(-\sin y +4 )\frac{\mathrm{d}y }{\mathrm{d} x}=(\cos x-9x^2)$

$\frac{\mathrm{d}y }{\mathrm{d} x}= \frac{(\cos x-9x^2)}{(4-\sin y )}$

Differentiate the following with respect to .

$3y^2+12x-6\cos x=-4y^3$

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-(\sin x +2)}{y +2 y^2}$

CORRECT

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{\sin x -2}{y +2 y^2}$

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-\sin x +2}{y +2 y^2}$

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-(\cos x +2)}{y +2 y^2}$

Explanation

The first step is to differentiate both sides with respect to :

$3y^2 \ \frac{\mathrm{d} }{\mathrm{d} x}+12x \ \frac{\mathrm{d} }{\mathrm{d} x}-6\cos x \ \frac{\mathrm{d} }{\mathrm{d} x}=-4y^3\frac{\mathrm{d} }{\mathrm{d} x}$