1. ## Integration/ Differentiation

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

attempt:

d/dθ( cosθ) = -sinθ
...

2. Originally Posted by mathcalculator22122
Use the derivative of cosθ to show that d/dθ ( secθ)= secθ tan θ

attempt:

d/dθ( cosθ) = -sinθ
...
let's change $\displaystyle \theta$ to $\displaystyle x$

$\displaystyle \sec x=\frac{1}{\cos x}$

Now use the quotient rule to take the derivative of both sides..

3. 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 $\displaystyle h(x)=cos(x)$

4. Just in case a picture helps...

... where

... 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!

5. 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

7. 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

8. Originally Posted by artvandalay11
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
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

9. thanks for helping