View Full Version : Implicit Differentiation - Trig

clannkelly

02-25-2012, 04:14 PM

I need some help with this. The example I'm stuck on is this:

d/dx of arcsin(5x+1)

The answer the book gives is this:

http://www.webassign.net/cgi-perl/symimage.cgi?expr=5%2F%28sqrt%28-25x%5E2%20-%2010x%29%29

I understand the basic concept of finding the derivative. My teacher keeps insisting that we have to multiply both numerator and denominator by the coefficient, but he can't seem to explain why in a way I understand. Can anyone help?

I need some help with this. The example I'm stuck on is this:

d/dx of arcsin(5x+1)

The answer the book gives is this:

http://www.webassign.net/cgi-perl/symimage.cgi?expr=5%2F(sqrt(-25x^2 - 10x))

There is nothing implicit about this.

The method is \dfrac{5}{\sqrt{1-(5x+1)^2}}.

HallsofIvy

02-25-2012, 04:51 PM

I think you may be misunderstanding what your teacher is saying. If you were getting a "common denominator" to add or subtract fractions, say, then you would need to multiply both numerator and denominator by the same thing so you get a different form of the same fraction.

If you have the fraction, say, \frac{3}{4}, and want to add it to \frac{2}{5}, you want to multiply both numerator and denominator by 5 so that you have \frac{15}{20} while multiplying both numerator and denominator of \frac{2}{5} by 4 to get \frac{8}{20}. Now they both have denominator 20 and so can be added. But you have only changed the "form" of the fractions, not the value-\frac{15}{20}= \frac{3}{4} and \frac{8}{20}= \frac{2}{5} so you are still adding the same numbers.

But if you are multiplying, say, 5 by a fraction, you are not just changing the form of a fraction you already have, you are multiplying to produce a new number. I am sure you have learned that \frac{a}{b}\times\frac{c}{d}= \frac{ac}{bd}. Well, 5= \frac{5}{1} so 5(\frac{a}{b})= \frac{5}{1}\frac{a}{b}= \frac{5a}{b}. Because you did not multiply both numerator and denominator by the same thing you do not have the same fraction now. But you shouldn't have the same fraction- you should have one 5 times as large.

Any calculus book will tel you that the derivative of arcsin(x) is \frac{1}{\sqrt{1- x^2}} so the derivative of arcsin(u), if u is a function of x, is, by the chain rule, \frac{1}{\sqrt{1- u^2}}\frac{du}{dx}. Here, u(x)= 5x+ 1 so that du/d= 5. The derivative of sin^{-1}(5x+10) is \frac{1}{\sqrt(1- (5x+1)^2}}(5)= \frac{5}{\sqrt{1- 25x^2- 10x+ 1}}= \frac{5}{\sqrt{-25x^2- 10x}}

clannkelly

02-25-2012, 05:16 PM

There is nothing implicit about this.

The method is \dfrac{5}{\sqrt{1-(5x+1)^2}}.

Well, it was in the chapter on implicit differentiation in my textbook. My professor doesn't really bother to teach us anything so I do a lot of guessing of terminology.

clannkelly

02-25-2012, 05:22 PM

I think you may be misunderstanding what your teacher is saying. If you were getting a "common denominator" to add or subtract fractions, say, then you would need to multiply both numerator and denominator by the same thing so you get a different form of the same fraction.

If you have the fraction, say, \frac{3}{4}, and want to add it to \frac{2}{5}, you want to multiply both numerator and denominator by 5 so that you have \frac{15}{20} while multiplying both numerator and denominator of \frac{2}{5} by 4 to get \frac{8}{20}. Now they both have denominator 20 and so can be added. But you have only changed the "form" of the fractions, not the value-\frac{15}{20}= \frac{3}{4} and \frac{8}{20}= \frac{2}{5} so you are still adding the same numbers.

But if you are multiplying, say, 5 by a fraction, you are not just changing the form of a fraction you already have, you are multiplying to produce a new number. I am sure you have learned that \frac{a}{b}\times\frac{c}{d}= \frac{ac}{bd}. Well, 5= \frac{5}{1} so 5(\frac{a}{b})= \frac{5}{1}\frac{a}{b}= \frac{5a}{b}. Because you did not multiply both numerator and denominator by the same thing you do not have the same fraction now. But you shouldn't have the same fraction- you should have one 5 times as large.

Any calculus book will tel you that the derivative of arcsin(x) is \frac{1}{\sqrt{1- x^2}} so the derivative of arcsin(u), if u is a function of x, is, by the chain rule, \frac{1}{\sqrt{1- u^2}}\frac{du}{dx}. Here, u(x)= 5x+ 1 so that du/d= 5. The derivative of sin^{-1}(5x+10) is \frac{1}{\sqrt(1- (5x+1)^2}}(5)= \frac{5}{\sqrt{1- 25x^2- 10x+ 1}}= \frac{5}{\sqrt{-25x^2- 10x}}

Let me see if I've got this right- because this is the chain rule, I'm supposed to multiply the derivative of the outer functions by the derivative of the inner functions. Is 5x an inner function? The derivative of 5x is 5, so is that why I'm supposed to multiply by 5?

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