Thread: Proving 2 derivatives equal to each other.

1. Proving 2 derivatives equal to each other.

f(x)=sec2x and g(x)=tan2x. Show that f'(x)=g'(x)

Well I didn't know how to do secant or tangent very well.
So I converted it into sine and cosine.
sec2x = 1/cos2x
Then I took the derivate of cosine
f'(x)=1/-sin2x

Then somehow I think I'm supposed to get g(x) into that as well?

Can someone explain it to me so I could understand it better.

P.S. Also does anyone know of a good website for trigonometri functions that convert sine, cosines, tagent, etc.
Like for example, like sinx/x = 1, or 1-cosx/cox=0, then there are a bunch of other rules. Anyone know the website of that? I used to have them in my notes, but then I lost the paper

2. Re: Proving 2 derivatives equal to each other.

Originally Posted by Chaim
f(x)=sec2x and g(x)=tan2x. Show that f'(x)=g'(x)

Well I didn't know how to do secant or tangent very well.
So I converted it into sine and cosine.
sec2x = 1/cos2x
Then I took the derivate of cosine
f'(x)=1/-sin2x

Then somehow I think I'm supposed to get g(x) into that as well?

Can someone explain it to me so I could understand it better.

P.S. Also does anyone know of a good website for trigonometri functions that convert sine, cosines, tagent, etc.
Like for example, like sinx/x = 1, or 1-cosx/cox=0, then there are a bunch of other rules. Anyone know the website of that? I used to have them in my notes, but then I lost the paper
I don't know of a web site of hand but bu the pythagorean identity we have that

$\displaystyle \tan^2(x)+1=\sec^2(x)$

So note that the two functions only differ by a constant.

Also your derivative is incorrect. You would need to use the quotient rule to take the derivative.

$\displaystyle f(x)=\frac{1}{\cos^2(x)} \implies f'(x)=\frac{0\cdot \cos^2(x)-1\cdot 1\cdot \sin(x)\cos(x)}{\cos^4(x)}=-\sec^2(x)\tan(x)$

3. Re: Proving 2 derivatives equal to each other.

you are expected to know the derivative of $\displaystyle \sec{x}$ and $\displaystyle \tan{x}$ ... look them up and/or derive them.

$\displaystyle f(x) = \sec^2{x} = (\sec{x})^2$

chain rule ...

$\displaystyle f'(x) = 2(\sec{x}) \cdot \sec{x}\tan{x}$

$\displaystyle g(x) = \tan^2{x} = (\tan{x})^2$

chain rule again ...

$\displaystyle g'(x) = 2(\tan{x}) \cdot \sec^2{x}$

4. Re: Proving 2 derivatives equal to each other.

Hello, Chaim!

Your differentiation is incorrect . . . way off!

$\displaystyle f(x)\:=\:\sec^2\!x\,\text{ and }\,g(x)\:=\:\tan^2\!x.\;\;\text{ Show that }f'(x)\:=\:g'(x)$

We have: .$\displaystyle f(x) \:= \: (\sec x)^2 \:=\:\frac{1}{(\cos x)^2} \:=\: (\cos x)^{-2}$

Then: .$\displaystyle f'(x) \:=\:\text{-}2(\cos x)^{-3}(\text{-}\sin x) \:=\:\boxed{\frac{2\sin x}{\cos^3x}}$

We have: .$\displaystyle g(x) \:=\: (\tan x)^2 \:=\: \frac{(\sin x)^2}{(\cos x)^2}$

Then: .$\displaystyle g'(x) \;=\;\frac{\cos^2\!x\cdot 2\sin x\cos x - \sin^2\!x\cdot 2\cos x(\text{-}\sin x)}{\cos^4\!x}$

n . . . . $\displaystyle g'(x) \;=\;\frac{2\sin x\cos^3\!x + 2\sin^3\!x\cos x}{\cos^4\!x}$

n . . . . $\displaystyle g'(x) \;=\;\frac{2\sin x\cos x\overbrace{(\cos^2\!x + \sin^2\!x)}^\text{This is 1}}{\cos^4\!x}$

n . . . . $\displaystyle g'(x) \;=\; \frac{2\sin x\cos x}{\cos^4\!x} \;=\; \boxed{\frac{2\sin x}{\cos^3\!x}}$