Implicit Differentiation

**Mindless** · January 18th 2011, 07:12 PM

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.

Thanks in advance.

**dwsmith** · January 18th 2011, 07:13 PM

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.

Thanks in advance.

What were your steps to arrive at cosh(y)?

**Mindless** · January 18th 2011, 07:18 PM

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

**dwsmith** · January 18th 2011, 07:19 PM

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.

**Mindless** · January 18th 2011, 07:26 PM

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

**dwsmith** · January 18th 2011, 07:34 PM

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?

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

**Mindless** · January 18th 2011, 07:38 PM

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).

**Chris L T521** · January 18th 2011, 07:59 PM

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.

Thanks in advance.

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

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

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

I hope this helps!

**Mindless** · January 19th 2011, 10:08 AM

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.

**dwsmith** · January 19th 2011, 11:38 AM

Originally Posted by Chris L T521

$\sqrt{1+\sinh^2y} = \sqrt{1+\left(\dfrac{4\sinh x-3}{4+3\sinh x}\right)^2}=\ldots$

Why is this?

**TheEmptySet** · January 19th 2011, 11:59 AM

Originally Posted by dwsmith

Why is this?

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

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

But in the OP we were given that

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

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