differentiation 4

• Nov 24th 2009, 10:39 PM
furor celtica
differentiation 4
given that siny= 2sinx, show that (dy/dx)^2 = 1 + 3sec^2y
by differentiating this equation with respect to x show that
d^2y/dx^2 = 3sec^2ytany
and hence that coty (d^2y/dx^2) - (dy/dx)^2 + 1 = 0
i havent gotten past the first problem:
differentiating siny=2sinx i get dy/dx= 2cosx/cosy
squaring this i get (dy/dx)^2 = 4cos^2/cos^2y; from here i use the first given statement and double angle identities to get a variety of results such as (4 - siny) x sec^2y but never 1+3sec^2y
can someone please show me exactly how to get to this result?
• Nov 25th 2009, 03:11 AM
HallsofIvy
Quote:

Originally Posted by furor celtica
given that siny= 2sinx, show that (dy/dx)^2 = 1 + 3sec^2y
by differentiating this equation with respect to x show that
d^2y/dx^2 = 3sec^2ytany
and hence that coty (d^2y/dx^2) - (dy/dx)^2 + 1 = 0
i havent gotten past the first problem:
differentiating siny=2sinx i get dy/dx= 2cosx/cosy
squaring this i get (dy/dx)^2 = 4cos^2/cos^2y; from here i use the first given statement and double angle identities to get a variety of results such as (4 - siny) x sec^2y but never 1+3sec^2y
can someone please show me exactly how to get to this result?

You have a good start, you have [tex](dy/dx)^2= \frac{4 cos^2(x)}{cos^2(y)}. Now, to replace that "cos(x)" with y you need to get back to "sin(x)". Use $cos^2(x)= 1- sin^2(x)$ to get $(dy/dx)^2= \frac{4(1- sin^2(x))}{cos^2(y)}= \frac{4- 4 sin^2(x)}{cos^2(y)}$ $= \frac{4- sin^2(y)}{cos^2(y)}= \frac{4}{cos^2(y)}- \frac{sin^2(y)}{cos^2(y)}$ $= 4sec^2(y)- tan^2(y)$.

Now use the fact that $tan^2(y)= sec^2(y)- 1$.
• Nov 25th 2009, 09:36 PM
furor celtica
"Now use the fact that http://www.mathhelpforum.com/math-he...efc334ca-1.gif."
i wasnt aware of this identity, can you show me how it is developed, either from basic principles or another basic identity?
• Nov 25th 2009, 10:22 PM
furor celtica
how am i supposed to solve the second problem?
i differentiated and got
3secytany/sqrt(3secy+1)
does this mean that i have to prove sqrt(3secy+1) = 1?
• Nov 26th 2009, 09:39 PM
Em Yeu Anh
Quote:

Originally Posted by furor celtica
"Now use the fact that http://www.mathhelpforum.com/math-he...efc334ca-1.gif."
i wasnt aware of this identity, can you show me how it is developed, either from basic principles or another basic identity?

You are aware of $sin^2x + cos^2x = 1$, I presume? It's a very common identity.
Divide all terms by $cos^2x$:

$\frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}$

$tan^2x + 1 = sec^2x$ which is equivalent to $tan^2x = sec^2x - 1$
• Nov 26th 2009, 11:40 PM
furor celtica
can you help me with the second problem?
• Jan 25th 2010, 02:44 AM
mr fantastic
Quote:

Originally Posted by furor celtica
can you help me with the second problem?

The way you formatted your original post makes it difficult to discern exactly what the second problem is, which would explain the lack of further replies.
• Jan 25th 2010, 06:09 AM
HallsofIvy
Quote:

Originally Posted by furor celtica
given that siny= 2sinx, show that (dy/dx)^2 = 1 + 3sec^2y

By "second problem" I presume you mean

Quote:

by differentiating this equation with respect to x show that
d^2y/dx^2 = 3sec^2ytany
and hence that coty (d^2y/dx^2) - (dy/dx)^2 + 1 = 0
Perhaps the reason you are not getting responses to that is that it is wrong!

Differentiating $(dy/dx)^2= 1+ 3sec^2 y$ we get $2(dy/dx)(d^2y/dx^2)= 6 sec y (sec y tan y)= 6 sec^2 y tan y$ so that $(dy/dx)(d^2y/dx^2)= 3 sec^2 y tan y$. But, since dy/dx is not 1, that is NOT $d^2y/dx^2= 3 sec^2 y tan y$.