# If x=a cos^3x and y=b sin^3x.find dy/dx?

If x=a cos^3x and y=b sin^3x.find dy/dx?

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$y ' = - \frac{b}{a} \tan x$

#### Explanation:

We have that:

$\frac{x}{a \cos x} = {\cos}^{2} x$

and:

$\frac{y}{b \sin x} = {\sin}^{2} x$

Then:

$\frac{x}{a \cos x} + \frac{y}{b \sin x} = 1$

Differentiate implicitly:

$\frac{a \cos x + a x \sin x}{{a}^{2} {\cos}^{2} x} + \frac{b y ' \sin x - b y \cos x}{{b}^{2} {\sin}^{2} x} = 0$

$\frac{1}{a \cos x} + \frac{x \sin x}{a {\cos}^{2} x} + \frac{y '}{b \sin x} - \frac{y \cos x}{b {\sin}^{2} x} = 0$

$\frac{y '}{b \sin x} = \frac{y \cos x}{b {\sin}^{2} x} - \frac{x \sin x}{a {\cos}^{2} x} - \frac{1}{a \cos x}$

$y ' = \frac{y \cos x}{\sin x} - \frac{b x {\sin}^{2} x}{a {\cos}^{2} x} - \frac{b \sin x}{a \cos x}$

Substitute now $x$ and $y$ from the original equations:

$y ' = \frac{b {\sin}^{3} x \cos x}{\sin x} - \frac{a b {\cos}^{3} x {\sin}^{2} x}{a {\cos}^{2} x} - \frac{b \sin x}{a \cos x}$

and simplify:

$y ' = \cancel{b {\sin}^{2} x \cos x} - \cancel{b \cos x {\sin}^{2} x} - \frac{b \sin x}{a \cos x}$

Then:

$y ' = - \frac{b}{a} \tan x$

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