derivative of an inverse trig function


by efekwulsemmay
Tags: derivative, function, inverse, trig
efekwulsemmay
efekwulsemmay is offline
#1
Feb12-10, 04:08 PM
P: 53
1. The problem statement, all variables and given/known data

[tex]y=sec^{-1}\frac{1}{t}, 0<t<1[/tex]

2. Relevant equations

[tex]\frac{d}{dx}sec^{-1} x= \frac{1}{\left|x\right|\cdot\sqrt{x^{2}-1}}[/tex]

3. The attempt at a solution
Basically to simplify things I used u substitution so I let u=1/t then du/dt=-1/t2 and I got:

[tex]y=sec^{-1}u\rightarrow y^{,}=\frac{1}{\left|u\right|\cdot\sqrt{u^{2}-1}}\cdot\frac{du}{dt}[/tex]

which,when I substituted for u, I got:

[tex]=\frac{1}{\left|\frac{1}{t}\right|\cdot\sqrt{\left(\frac{1}{t}\right)^{ 2}-1}}\cdot\frac{-1}{t^{2}}[/tex]

which works out as:

[tex]=\frac{-1}{\left|\frac{1}{t}\right|\cdot t^{2}\cdot\sqrt{\frac{1^{2}}{t^{2}}-1}}[/tex]

and:

[tex]=\frac{-1}{t\cdot\sqrt{1-t^{2}}}[/tex]

however, the answer as per the back of the book is:

[tex]\frac{-1}{\sqrt{1-t^{2}}}[/tex]

what did I do wrong???
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rasmhop
rasmhop is offline
#2
Feb12-10, 04:19 PM
P: 418
The step,
[tex]\frac{-1}{\left|\frac{1}{t}\right|\cdot t^{2}\cdot\sqrt{\frac{1^{2}}{t^{2}}-1}} = \frac{-1}{t\cdot\sqrt{1-t^{2}}}[/tex]
is incorrect. Remember
[tex]a\sqrt{b} = \sqrt{a^2b}[/tex]
so you have,
[tex]t\sqrt{1/t^2-1} = \sqrt{1-t^2}[/tex]
efekwulsemmay
efekwulsemmay is offline
#3
Feb12-10, 04:24 PM
P: 53
ooooooooh ok. that makes more sense. thats clever hehehe

thank you so much

Gib Z
Gib Z is offline
#4
Feb12-10, 05:21 PM
HW Helper
Gib Z's Avatar
P: 3,353

derivative of an inverse trig function


For next time, you may remember that since [tex]\cos x = \frac{1}{\sec x}[/tex], it is true that [tex]\sec^{-1} \left(\frac{1}{x}\right) = \cos^{-1} x[/tex] and so the derivative is standard.


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