# Implicit Differentiation

• Jan 18th 2011, 07:12 PM
Mindless
Implicit Differentiation
Hello there. I'm new to this forum so I apologise in advance if what I have posted seems a little unclear to you. I just need a little help with the following question.

If, sinhy = (4sinhx - 3) / (4 + 3sinhx)

show that,

dy/dx = -5 / (4 + 3sinhx)

My final answer contains coshy, and I do not know where to go from there to get the above.

• Jan 18th 2011, 07:13 PM
dwsmith
Quote:

Originally Posted by Mindless
Hello there. I'm new to this forum so I apologise in advance if what I have posted seems a little unclear to you. I just need a little help with the following question.

If, sinhy = (4sinhx - 3) / (4 + 3sinhx)

show that,

dy/dx = -5 / (4 + 3sinhx)

My final answer contains coshy, and I do not know where to go from there to get the above.

What were your steps to arrive at cosh(y)?
• Jan 18th 2011, 07:18 PM
Mindless
Your really going to make me type all of that? lol. I basically used the quotient rule which gave me the final result as follows,

25cosh(x) / cosh(y) (4 + 3sinh(x))^2
• Jan 18th 2011, 07:19 PM
dwsmith
Quote:

Originally Posted by Mindless
Your really going to make me type all of that? lol. I basically used the quotient rule which gave me the final result as follows,

25cosh(x) / cosh(y) (4 + 3sinh(x))^2

It is forum requirement. Plus, it is easier to find a mistake then to type up the whole solution.
• Jan 18th 2011, 07:26 PM
Mindless
That's a very valid point. Okay so these are the steps I took (keeping in mind I used the quotient rule).

differentiating w.r.t x we get the following,

cosh(y) dy/dx = [(4 + 3sinh(x))(4cosh(x)) - (4sinh(x) - 3)(3cosh(x)] / (4 + 3sinh(x))^2

open up the numerator to get,

cosh(y) dy/dx = 25cosh(x) / (4 + 3sinh(x))^2

thus,

dy/dx = 25cosh(x) /cosh(y) (4 + 3sinh(x))^2
• Jan 18th 2011, 07:34 PM
dwsmith
Quote:

Originally Posted by Mindless
That's a very valid point. Okay so this is the steps I took (keeping in mind I used the quotient rule).

differentiating w.r.t x we get the following,

cosh(y) dy/dx = [(4 + 3sinh(x))(4cosh(x)) - (4sinh(x) - 3)(3cosh(x)] / (4 + 3sinh(x))^2

open up the numerator to get,

cosh(y) dy/dx = 25cosh(x) / (4 + 3sinh(x))^2

thus,

dy/dx = 25cosh(x) /cosh(y) (4 + 3sinh(x))^2

Have you tried converting the hyperbolics to exponentials and then simplifying?

$\displaystyle \displaystyle\cosh{x}=\frac{e^x+e^{-x}}{2} \ \ \ \ \ \sinh{x}=\frac{e^x-e^{-x}}{2}$
• Jan 18th 2011, 07:38 PM
Mindless
No, but the cosh(y) is posing a big problem. I have no idea how to get rid of that, so that I am left with a function which only depends on the variable (x).
• Jan 18th 2011, 07:59 PM
Chris L T521
Quote:

Originally Posted by Mindless
Hello there. I'm new to this forum so I apologise in advance if what I have posted seems a little unclear to you. I just need a little help with the following question.

If, sinhy = (4sinhx - 3) / (4 + 3sinhx)

show that,

dy/dx = -5 / (4 + 3sinhx)

My final answer contains coshy, and I do not know where to go from there to get the above.

Note that $\displaystyle \dfrac{\,dy}{\,dx}=\dfrac{25\cosh x}{(\cosh y)(4+3\sinh x)^2}$, as you correctly said.

Now, recall the identity $\displaystyle \cosh^2y-\sinh^2y=1$. So in your case $\displaystyle \cosh y = \sqrt{1+\sinh^2y} = \sqrt{1+\left(\dfrac{4\sinh x-3}{4+3\sinh x}\right)^2}=\ldots$

If you get $\displaystyle \cosh y$ in terms of x, you should be able to get the answer. However, when I did this, I got $\displaystyle \dfrac{\,dy}{\,dx}=\dfrac{5}{4+3\sinh x}$ -- not the negative value you provided.

I hope this helps!
• Jan 19th 2011, 10:08 AM
Mindless
That was really helpful! yes your right Chris, I also end up with a positive value. Maybe some further fiddling around could bring out the negative five, or maybe the book is incorrect? :P. Anyway, thank you very much for your help. Much appreciated.
• Jan 19th 2011, 11:38 AM
dwsmith
Quote:

Originally Posted by Chris L T521
$\displaystyle \sqrt{1+\sinh^2y} = \sqrt{1+\left(\dfrac{4\sinh x-3}{4+3\sinh x}\right)^2}=\ldots$

Why is this?
• Jan 19th 2011, 11:59 AM
TheEmptySet
Quote:

Originally Posted by dwsmith
Why is this?

$\displaystyle \cosh^2(y)-\sinh^2(y)=1$ is an identity for hyperbolic trig functions this gives

$\displaystyle \cosh^2(y)=1+\sinh^2(y) \implies \cosh(y)=\sqrt{1+\sinh^2(y) }$

But in the OP we were given that

$\displaystyle \sinh(y)=\frac{4\sinh(x)-3}{4+3\sinh(x)}$