# Math Help - differential

1. ## differential

i need general instructions for this one...

thanks!

2. Hello, Phoebe!

Let: . $y\:=\:\cosh(x^2 - 3x + 1)$ . where: . $\cosh t \:= \:\frac{e^t - e^{-t}}{2}$

Find: . $\frac{d^2y}{dx^2}$

If we are allowed to use these differentiation formula, it's easier.
. . $\frac{d}{dx}(\sinh x) \:=\:\cosh x\qquad\qquad\frac{d}{dx}(\cosh x) \:=\:\sinh x$

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We have: . $y \:=\:\cosh(x^2 - 3x + 1)$

Then: . $\frac{dy}{dx}\:=\:\cosh(x^2-3x + 1)\cdot(2x - 3)$

And: . $\frac{d^2y}{dx^2}\:=\:\sinh(x^2-3x+1)\!\cdot\!2 \:+ \:(2x-3)\!\cdot\!\cosh(x^2-3x+1)\!\cdot\!(2x - 3)$

. . . . $\boxed{\frac{d^2y}{dx^2} \:= \:2\sinh(x^2-3x+1) + (2x-3)^2\cosh(x^2-3x+1)}$

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They may expect us to use the Chain Rule on:

. . $y \:=\:\frac{1}{2}\left[e^{(x^2-3x+1)} + e^{-(x^2-3x+1)}\right]$ . . . ack!

3. Thanks!

now i understand that d^2y/dX^2 is just the second derivative...

i think they expect us to use the exponent (e) with the chain rule... i can do it now.