Cosh^-1(x)

Original post by megafobia

1/(x^(2)-1)^(-1/2)

That is the formula of the derivative.

Then I simply put x into that squared, which was (5x+3)^(-1/2)

Do you know the Chain Rule?

Reply 7

Original post by Mr M

Do you know the Chain Rule?

Yeah

Reply 8

Original post by megafobia

Yeah

u=\sqrt{5x+3}

y=\cosh^{-1} u

\frac{du}{dx}=...

\frac{dy}{du}=...

\frac{dy}{dx}=\frac{dy}{du} \times \frac{du}{dx}

(edited 11 years ago)

Reply 9

SubAtomic

You can also use this formula

\displaystyle (f^{-1})^\prime(x) =\frac {1}{f^\prime(f^{-1}(x))}

(edited 11 years ago)

Reply 10

To be honest it is just as quick to implicitly differentiate

\cosh y = \sqrt{5x+3}

\sinh y \frac{dy}{dx}=\frac{5}{2\sqrt{5x+3}}

Now

\sinh^2 y = \cosh^2 y - 1

...

Reply 11

Original post by Mr M

To be honest it is just as quick to implicitly differentiate

\cosh y = \sqrt{5x+3}

\sinh y \frac{dy}{dx}=\frac{5}{2\sqrt{5x+3}}

Now

\sinh^2 y = \cosh^2 y - 1

...

I tried this and got

(edited 11 years ago)

Reply 12

Original post by megafobia

I tried this and got

That isn't right either.

What's your mathematical background?

Reply 13

Original post by Mr M

That isn't right either.

What's your mathematical background?

Undergraduate 1st year (with a level maths)

Reply 14

As you are really struggling here's a worked solution:

Spoiler

Try to understand each step.

Reply 15

Original post by megafobia

Undergraduate 1st year (with a level maths)

Undergraduate of what subject?

Reply 16

Original post by Mr M

Undergraduate of what subject?

maths

Reply 17

Original post by Mr M

As you are really struggling here's a worked solution:

Spoiler

Try to understand each step.

I get everything you've done. Are sure that it is right? I've tried putting it into this uni thing and it says that it is wrong.

Reply 18