Find the Derivative of y=sec(ax) using defition

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SUMMARY

The derivative of y=sec(ax) can be derived using the limit definition and approximations. The correct derivative is a sec(ax)tan(ax), which can be confirmed through the limit process involving the approximations of sin(ah) and cos(ah). Key steps include applying the limit as h approaches 0 and utilizing trigonometric identities to simplify the expression. The discussion highlights the importance of correctly organizing terms in the limit to facilitate the derivation.

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  • Knowledge of the derivative definition
  • Experience with Taylor series approximations
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Homework Statement


I need to show the derivative of y=sec(ax) using the definition and aproximations

(Hopefully I'll type this so its understandable what I've done)

Homework Equations



I realize the answer is sin(ax)/cos2(ax)
I also can do get the answer using the quotient rule but we're supposed to do it using aprox and definitons


The Attempt at a Solution



Lim Sec(a(x+h) - sex(ax)
h->0 h

sec(ax+ah) = 1/cos(ax+ah)
cos(ax+ah) = cos(ax)cos(ah)-sin(ax)sin(ah)

ah as h->0 is small so we can approximate sin(ah) and cos(ah)
Linear aprox of sin(ah) = ah + ah3/6
Linear aprox of cos(ah) = 1 + ah2/2

this gives us
cos(ax)(1 + ah2/2) - sin(ax)(ah + ah3/6

cos(ax) + cos(ax)ah2/2 -ah*sin(ax) - sin(ax)ah3/6

plug back into limit
(cos(ax) - cos(ax) kills that)
lim 1 / cos(ax)ah2/2 -ah*sin(ax) - sin(ax)ah3/6 / h
h->0

this is where I'm stuck. I know there some identity or math trickery that's needed but I have no idea where to proceed.
 
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cpscdave said:

Homework Statement


I need to show the derivative of y=sec(ax) using the definition and aproximations

(Hopefully I'll type this so its understandable what I've done)

Homework Equations



I realize the answer is sin(ax)/cos2(ax)
Actually, it's a times the above, or a sec(ax)tan(ax).
cpscdave said:
I also can do get the answer using the quotient rule but we're supposed to do it using aprox and definitons


The Attempt at a Solution



Lim Sec(a(x+h) - sex(ax)
h->0 h

sec(ax+ah) = 1/cos(ax+ah)
cos(ax+ah) = cos(ax)cos(ah)-sin(ax)sin(ah)

ah as h->0 is small so we can approximate sin(ah) and cos(ah)
Linear aprox of sin(ah) = ah + ah3/6
Linear aprox of cos(ah) = 1 + ah2/2

this gives us
cos(ax)(1 + ah2/2) - sin(ax)(ah + ah3/6

cos(ax) + cos(ax)ah2/2 -ah*sin(ax) - sin(ax)ah3/6

plug back into limit
(cos(ax) - cos(ax) kills that)
lim 1 / cos(ax)ah2/2 -ah*sin(ax) - sin(ax)ah3/6 / h
h->0
This is very hard to read. You have four / signs indicating division, so it's difficult to determine what's in the numerator and what's in the denominator.
cpscdave said:
this is where I'm stuck. I know there some identity or math trickery that's needed but I have no idea where to proceed.

\frac{1}{h}\cdot \frac{cos(ax) - cos(ax)cos(ah) + sin(ax)sin(ah)}{(cos(ax)cos(ah) - sin(ax)sin(ah))cos(ax)}

To see my LaTeX script, click on the expression above.

Next, write this as two fractions, using the first two terms in the numerator for the first fraction, and the third term for the other fraction.

Another approach besides approximating sin(ah) and cos(ah) is to work with the limits
\lim_{h \to 0}\frac{sin(ah)}{h} = a
and
\lim_{h \to 0}\frac{1 - cos(ah)}{h} = 0
 
Perfect that clued me in exactly where I was going wrong.

Never occurred to me in the first step to put sec(ax+ah) and sec(ax) over a common denominator.

Thanks :D
 

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