implicit differentiation

R

red and white kop!

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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?
 
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
(dy/dx)^2 = 4cos^2x/cos^2y
Click to expand...

4cos^2x/cos^2y
= 4(1-sin^2x)/cos^2y
= (4 – 4sin^2x)/cos^2y
= [4 – (2sinx)^2]/cos^2y
= [4 – sin^2y]/cos^2y
= [3 + (1) - sin^2y]/cos^2y
= [3 + (sin^2y + cos^2y) - sin^2y]/cos^2y
= [3 + cos^2y]/cos^2y
= 3/cos^2y + cos^2y/ cos^2y
= 1 + 3/cos^2y
= 1 + 3sec^2y

Hope that helps. :)
 
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?
 
dy/dx= 2cosx/cosy

i differentiated and got
3secytany/sqrt(3secy+1)
Click to expand...

?
Please show us your work so we can see where you need help.
 
im differentiating from the last equation we got, i.e. (dy/dx)^2 in terms of y, as we are supposed to find d^2y/dx^2 in terms of y, as explained.
im working on the second problem mentioned in the opening post, btw
thanks for your help
 
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

i differentiated and got
3secytany/sqrt(3secy+1)
Click to expand...

I asked you to show your work because your second derivative is incorrect. I was hoping you would catch your mistakes if you went through it again. You are missing a squared term, and you forgot dy/dx on the end.

d^2y/dx^2 = [3sec^2ytany/sqrt(3secy+1)](dy/dx)
d^2y/dx^2 = [3sec^2ytany/sqrt(3secy+1)](sqrt(3secy+1))
d^2y/dx^2 = 3sec^2ytany

Cheers!
 
what? can you please explain this, i dont understand how you got that extra dy/dx! ive been mulling over this exercise for days
 
\(\displaystyle We \ got \ this \ far, \ correct? \ (y')^{2} \ = \ 1+3sec^{2}(y)\)

\(\displaystyle Hence, \ chain \ rule, \ 2(y')(y") \ = \ 0+6sec(y)[sec(y)tan(y)]y'\)

\(\displaystyle Ergo, \ y" \ = \ 3sec^{2}(y)tan(y) \ QED\)
 
\(\displaystyle For \ those \ of \ you \ from \ Rio \ Linda:\)

\(\displaystyle \bigg(\frac{dy}{dx}\bigg)^{2} \ = \ 1+3sec^{2}(y), \ now \ taking \ the \ derivative \ of \ both \ sides \ with \ respect \ to\)

\(\displaystyle \ x, \ yields: \ 2\bigg(\frac{dy}{dx}\bigg)\bigg(\frac{d^{2}y}{dx^{2}}\bigg) \ =0+ \ 6sec(y)[sec(y)tan(y)]\bigg(\frac{dy}{dx}\bigg)\)

\(\displaystyle This \ gives \ \frac{d^{2}y}{dx^{2}} \ = \ 3sec^{2}(y)tan(y)\)

\(\displaystyle Note: \ D_x[sin(x)]^{2} \ = \ 2[sin(x)][cos(x)]\)

\(\displaystyle Hence, \ D_x\bigg(\frac{dy}{dx}\bigg)^{2} \ = \ 2\bigg(\frac{dy}{dx}\bigg)\bigg(\frac{d^{2}y}{dx^{2}}\bigg)\)

\(\displaystyle Ergo, knowing \ the \ properties \ of \ derivatives, \ and \ the \ chain \ rule, \ \ will \ expedite \ matters \ and \ save\)

\(\displaystyle \ a \ lot \ of \ grunt \ work.\)
 
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