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differentiation (hyperbolic ?)

Could anyone help me with question 4, I've been on it for an hour and got no clue how to do it.

IMG_20131009_154440.jpg502.0KB

Original post by ibrahim541

Could anyone help me with question 4, I've been on it for an hour and got no clue how to do it.

\sinh (x) = \frac{1}{2} \left( e^x - e^{-x} \right)

What similarities can you see between

y

and

\sinh (x) \ ?

(edited 10 years ago)

OP

Original post by Khallil

\sinh (x) = \frac{1}{2} \left( e^x - e^{-x} \right)

What similarities can you see between

y

and

\sinh (x) \ ?

Y(x) = 2sinh(x) ?

Original post by ibrahim541

Y(x) = 2sinh(x) ?

Exactly

:smile:

Do you know how to differentiate hyperbolic functions?

OP

Original post by Khallil

Exactly

:smile:

Do you know how to differentiate hyperbolic functions?

I do and after I differentiate it I get dy/dx=2cosh(x) but it wants it in terms of y and I've got no idea how to rearrange the original formula to get that

There's a useful identity:

\cosh^2 (x) - \sinh^2 (x) = 1

Original post by ibrahim541

I do and after I differentiate it I get dy/dx=2cosh(x) but it wants it in terms of y and I've got no idea how to rearrange the original formula to get that

Ah I see. Have you learnt Osbourne's rule yet?

It states that to convert any trigonometric identity to a hyperbolic identity, you must change all cos terms into cosh and replace any product of two sin terms with -sinh^2 :

\cos \to \cosh

\sin \to \sinh

\sin \times \sin \to -\sinh^2

For example:

\cos^2 x + \sin^2 x = 1 \longrightarrow \cosh^2 x - \sinh^2 x = 1

\sin^2 x

is the product of two

\sin x

terms so it is replaced by

-\sinh^2 x

(edited 10 years ago)

OP

Ahh okay, that helps a lot. so I got dx/dy=1/2(y^2+1)^0.5 can anyone clarify if this is right ?

Original post by ibrahim541

...

y = 2\sinh x \ \Rightarrow \ \dfrac{dy}{dx} = 2\cosh x

\cosh^2 x = 1 + \sinh^2x \ \Rightarrow \ \cosh x = \sqrt{1+\underbrace{\sinh^2x}}

Use the fact that

\ \sinh x = \frac{y}{2} \

to simplify the expression for

\cosh x

(edited 10 years ago)

OP

Original post by Khallil

y = 2\sinh x \ \Rightarrow \ \dfrac{dy}{dx} = 2\cosh x

\cosh^2 x = 1 + \sinh^2x \ \Rightarrow \ \cosh x = \sqrt{1+\sinh^2x}

Use the fact that

\ \sinh x = \frac{y}{2} \

to simplify the expression for

\cosh x

Ah so its dx/dy=1/2(1+(y/2)^0.5)^0.5, thanks you so much for your help

Original post by ibrahim541

Ah so its dx/dy=1/2(1+(y/2)^0.5)^0.5, thanks you so much for your help

I think you mean:

\dfrac{dy}{dx} = 2 \sqrt{ 1 + \left( \frac{y}{2} \right)^2 }

(edited 10 years ago)

OP

Original post by Khallil

I think you mean:

\dfrac{dy}{dx} = 2 \sqrt{ 1 + \left( \frac{y}{2} \right)^2 }

Okay got it, thanks again

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