Use the derivative of cosθ to show that d/dθ ( secθ)= secθ tan θ

attempt:

d/dθ( cosθ) = -sinθ

...

Printable View

- Nov 6th 2009, 08:33 AMmathcalculator22122Integration/ Differentiation
Use the derivative of cosθ to show that d/dθ ( secθ)= secθ tan θ

attempt:

d/dθ( cosθ) = -sinθ

... - Nov 6th 2009, 08:39 AMartvandalay11
- Nov 6th 2009, 08:41 AMKeithfert488
The best way to go about proving that is to use the quotient rule.

If $\displaystyle f(x)=\frac{g(x)}{h(x)}$, then $\displaystyle f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{h(x)^2}$

Plug in $\displaystyle g(x)=1$ and $\displaystyle h(x)=cos(x)$ - Nov 6th 2009, 08:47 AMtom@ballooncalculus
Just in case a picture helps...

http://www.ballooncalculus.org/asy/secDiff.png

... where

http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which subject to the chain rule).

__________________________________________

Don't integrate - balloontegrate!

Balloon Calculus: Gallery

Balloon Calculus Drawing with LaTeX and Asymptote! - Nov 6th 2009, 08:50 AMartvandalay11
that isn't quicker at all, because then you have to bring the powers back down into the denominator to realize you actually have sec and tan

- Nov 6th 2009, 09:13 AMtom@ballooncalculus
Prettier though, surely! And how about this?

http://www.ballooncalculus.org/asy/diffChain/secAlt.png - Nov 6th 2009, 11:43 AMartvandalay11
I'm not saying it isnt valid or that you won't get the correct answer, but I can apply the chain rule without drawing circles and lines, and as such, it will be quicker

- Nov 6th 2009, 12:10 PMtom@ballooncalculus
Yes, and it won't create that 'powers' problem, will it?

But now the goal posts have moved, let's not bicker... I may have provoked you in the first place by daring to introduce my pic with 'quicker by balloon!', which I edited out within the minute because I remembered that this is only true for quotients already having a power in the denominator - see http://www.ballooncalculus.org/examp...ence.html#quot if interested - hope so!

Cheers

Tom - Nov 6th 2009, 04:34 PMmathcalculator22122
thanks for helping(Hi)