1. ## 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.

2. 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)?

3. 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

4. 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.

5. 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

6. 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?

$\displaystyle \displaystyle\cosh{x}=\frac{e^x+e^{-x}}{2} \ \ \ \ \ \sinh{x}=\frac{e^x-e^{-x}}{2}$

7. 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).

8. 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!

9. 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.

10. 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?

11. 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

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

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### If (sinhx-3)/(4 3sinhx) show that dy/DX=-5/(4 3sinhx)

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