Quick derivative question!Watch

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#1
Hi, I've got a quick question:

What is the derivative of sec^2(x) [sec squared x]

Thanks!
0
13 years ago
#2
Using the product rule you differentiate the outside which is 2secx and then times the differential of secx which is secxtanx so the answer is 2sec^2xtanx.
0
13 years ago
#3
(Original post by rockiee)
Hi, I've got a quick question:

What is the derivative of sec^2(x) [sec squared x]

Thanks!
Think u have to use the chain rule.

y=u^2 u=secx
dy/du =2u du/dx=tanxsecx
dy/dx= dy/du x du/dx (dy/du multiplied by du/dx)
dy/dx = 2u times tanxsecx substitute u=secx

dy/dx= 2secx tanxsecx = 2tanxsec^2x
Hope thats right.
0
13 years ago
#4
Beat me to it, but I've added it as example 4 here:

http://wikinotes.cdpuk.net/wiki/The_chain_rule

with a little more explanation

-Chris
0
13 years ago
#5
Doh! I meant the chain rule
0
13 years ago
#6
(Original post by rockiee)
Hi, I've got a quick question:

What is the derivative of sec^2(x) [sec squared x]

Thanks!
For some reason I feel you meant the integral of sec^2(x) ?

If so, its quite hard to do, and you would just differentiate tanx to show its the integral of sec^2(x)
0
#7
Thanks everyone! I also managed to get the answer through using the trig identity of sec^2(x) i.e. 1+tan^2(x) and I got the same answer as above.

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