Differentiation (1 Viewer)

chicktracts · Oct 9, 2009

What is the differentiation of cosec(x) and sec(x)?

tom.evans.15 · Oct 9, 2009

I would use the identity $\frac{d}{dx}cosec(x)=\frac{d}{dx}\frac{1}{\sin(x)}$ and then do $\frac{d}{dx}(\sin(x))^{-1}$ and use the chain rule to work it out...

Similiarly I would use the identity $sec(x)=\frac{1}{\cos(x)}$ to do the same
hope i helped

Drongoski · Oct 9, 2009

$\frac{\mathrm{d} }{\mathrm{d} x}(csc x)\ =\ \frac{\mathrm{d} }{\mathrm{d} x} (sinx)^{-1}\ =\ -\frac{1}{sin^{2}x}\cdot cos\ x\ =\ -\frac{1}{sinx}\cdot \frac{cosx}{sinx}\ =\ -cscx\cdot cotx\\ \\ \\ \frac{\mathrm{d} secx}{\mathrm{d} x}\ =\ \frac{\mathrm{d} (cosx)^{-1}}{\mathrm{d} x}\ =\ -\frac{1}{cos^{2}x}\cdot -sinx\ =\ \frac{1}{cosx}\cdot \frac{sinx}{cosx}\ =\ secx\cdot tanx$

Just corrected a very basic error pointed out by Mr Evans

tom.evans.15 · Oct 9, 2009

Just dont get confused between $\sin^{-1}(x)$ and $\frac{1}{\sin(x)}$ , they mean completely different things, $\sin^{-1}(x)$ is the inverse function, and $\frac{1}{\sin(x)}$ is the reciprocal of the function

cutemouse · Oct 9, 2009

I think d/dx (sec x) = secx tanx is given in the HSC paper.

khorne · Oct 9, 2009

It is...

Integral of sec (ax) tan (ax) dx = 1/a sec(ax)

lychnobity · Oct 9, 2009

cutemouse · Oct 9, 2009

khorne said:
It is...

Integral of sec (ax) tan (ax) dx = 1/a sec(ax)

Basically what I said

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