Help: understanding partial derivative

We are given $x = r\cos \theta$ and $y = r\sin \theta$. Find partial derivative of 'r' w.r.t 'x'.
I tried solving this way
$r = x\sec \theta$ .
$\frac{dr}{dx}$ = $\sec \theta$ . [ I have used the symbol 'd' for partial derivative].
Also,squaring and adding both equations in question we get,
$r^2 = x^2 + y^2$ .
$2r\frac{dr}{dx} = 2x$ .
$\frac{dr}{dx} = \frac{x}{r}$ .
$\frac{dr}{dx} = \frac{r\cos \theta}{r}$ [As given, $x=r\cos \theta$ ].
= $\cos \theta$ .
I got two different answers I am confused where am i wrong?

#Calculus

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Comments

Do you know what a partial derivative is?

Did you apply the product rule correctly?

Hint: Does $\theta$ depend on $x$ ?

Calvin Lin Staff - 4 years, 5 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$