Author Topic: Differentiation Problem (Read 809 times) Tweet Share

ice_blockie · « **on:** January 27, 2008, 02:13:46 pm »

I was working through Essential Chapter 9, and I came to this question:

EX 9E Q5: If $y=(x+\sqrt {x^2+1})^2$ , show $\frac{dy}{dx}=\frac {2y}{\sqrt {x^2+1}}$

Tis ugly, or I just can't see the answer. I worked through it so many times

Can anyone show me how to do this? Thanks!

Mao · « **Reply #1 on:** January 27, 2008, 02:25:52 pm »

mmm

MAJOR EDIT: thanks to dcc.... lol u bastard

chain rule:

$let \ u = x+ \sqrt{x^2+1}$ <----- inner function

$\frac{du}{dx} = 1 + \frac{d}{dx} \sqrt{x^2+1}$

$\frac{du}{dx} = 1+ \frac{x}{\sqrt{x^2+1}}$

$\frac{du}{dx} = \frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}}$

$y = u^2$ <--------- outer function

$\frac{dy}{du} = 2u$

$\frac{dy}{dx} = \frac{du}{dx} \cdot \frac{dy}{du}$

$\frac{dy}{dx} = \frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}} \cdot 2(x+ \sqrt{x^2+1})$

$\frac{dy}{dx} = 2 \cdot \frac{(x + \sqrt{x^2+1})^2}{\sqrt{x^2+1}}$

but since $y=(x+\sqrt {x^2+1})^2$

$\frac{dy}{dx} = 2 \cdot \frac{y}{\sqrt{x^2+1}}$

therefore

$\frac{dy}{dx}=\frac {2y}{\sqrt {x^2+1}}$

QED

note: in reality there's no need to lay out all these workings (which are done step by step)
you would probably do most of it mentally, it should take about 3/4 lines

Ahmad · « **Reply #2 on:** January 27, 2008, 02:36:01 pm »

Shortest way I can think of that you don't need to know for methods:
$y = e^{2\cdot \mbox{sinh^{-1}}(x)}$
$\frac{dy}{dx} = \frac{d}{dx} (2 \mbox{sinh^{-1}(x)}) \cdot e^{2\cdot \mbox{sinh^{-1}}(x)}$

$= \frac{2y}{\sqrt{x^2 + 1}}$

midas_touch · « **Reply #3 on:** January 27, 2008, 04:33:45 pm »

I like Ahmad's method better

.

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