1. ## Differentiation y w.r.t.x

x^-1 + y^-1 = e^y

My textbook has a pretty rubbish explanation of how to differentiate y, from what i gather its like x but you stick a dy/dx after it and use the product rule for xy terms, can anyone do the above question with a slightly better explanation than ive given?!
Thanks!

2. Originally Posted by darksupernova
x^-1 + y^-1 = e^y

My textbook has a pretty rubbish explanation of how to differentiate y, from what i gather its like x but you stick a dy/dx after it and use the product rule for xy terms, can anyone do the above question with a slightly better explanation than ive given?!
Thanks!
Hi darksupernova,

i guess you mean when you differentiate both sides with respect to x,
and y is a function of x.

Then you use $\displaystyle \frac{d}{dx}f(y)=\frac{dy}{dy}\frac{d}{dx}f(y)=\fr ac{dy}{dx}\frac{d}{dy}f(y)$

You can just think of swopping denominators, as in

$\displaystyle \left(\frac{10}{3}\right)\frac{6}{5}=\frac{60}{15} =\left(\frac{10}{5}\right)\frac{6}{3}$

It's called the chain rule

$\displaystyle \frac{d}{dx}x^{-1}+\frac{dy}{dx}\frac{d}{dy}y^{-1}=\frac{dy}{dx}\frac{d}{dy}e^y$

and continue

3. Thankyou, its a bit odd to get my head round but thats probably because ive never done it!

4. Hi darksupernova,

i forgot to mention,

you can differentiate y with respect to y,
we can differentiate x with respect to x.

when we have $\displaystyle \frac{d}{dx}f(y)$

and we don't have y expressed in terms of x,
we've got a dx where we'd prefer to have a dy

We can multiply any value by 1 without changing it,
and there are an infinite number of ways to write 1.

We get dy where we want it by multiplying by $\displaystyle \frac{dy}{dy}$
and swopping denominators.

5. okay i get that part, but this question has thrown me a little...

Find \frac{dy}{dx} as a function of x if y^2 = 2x + 1.

How do i make dy/dx a duction of x?

6. Originally Posted by darksupernova
okay i get that part, but this question has thrown me a little...

Find \frac{dy}{dx} as a function of x if y^2 = 2x + 1.

How do i make dy/dx a duction of x?
y^2 = 2x + 1
Differentiating this you get
2y*dy/dx = 2. Or
dy/dx = 1/y = 1/sqrt(2x + 1)

7. Originally Posted by darksupernova
okay i get that part, but this question has thrown me a little...

Find $\displaystyle \frac{dy}{dx}$ as a function of x if y^2 = 2x + 1.

How do i make dy/dx a function of x?
$\displaystyle y^2=2x+1$

$\displaystyle \frac{d}{dx}y^2=\frac{d}{dx}(2x+1)=2$

$\displaystyle \frac{dy}{dx}\frac{d}{dy}y^2=2$

now $\displaystyle y^2$ can be differentiated wrt y

$\displaystyle \frac{dy}{dx}2y=2$

$\displaystyle \frac{dy}{dx}=\frac{2}{2y}=\frac{1}{y}$

You get it by differentiating the x terms wrt x and the y terms wrt y,

then dividing by the multiplier of $\displaystyle \frac{dy}{dx}$

8. i differentiated it correctly but i didnt substitue y^2 = 2x + 1 back in :S is that because it says as a function of x?

9. Originally Posted by darksupernova
i differentiated it correctly but i didnt substitue y^2 = 2x + 1 back in :S is that because it says as a function of x?
yes, as a function of x, then continue as sa-ri-ga-ma showed.

$\displaystyle y^2=2x+1\ \Rightarrow\ y=\pm\sqrt{2x+1}$

Initially, you could also write

$\displaystyle y=\pm\sqrt{2x+1}$

and calculate

$\displaystyle \frac{dy}{dx}=\pm\frac{d}{dx}(2x+1)^{0.5}=\pm(2)(0 .5)(2x+1)^{-0.5}$

10. It should be
(0.5)(2)(2x + 1)^-0.5

11. yeah i got it now, thanks guys, it was just the fact i hadnt substituted in the value for y. Got it now, thank you both!

### differentiate of ysiny

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