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?

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TopNewestDo you know what a partial derivative is?

Did you apply the product rule correctly?

Hint:Does \( \theta \) depend on \(x\)? – Calvin Lin Staff · 8 months, 2 weeks agoLog in to reply