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Differentiation Help?

Hi all

As per my screenshot, how do I differentiate from fx to fxx?

Thanks in advance!

Reply 1
I cannot load the picture
Original post by Texcoco95
Hi all

As per my screenshot, how do I differentiate from fx to fxx?

[IMAGE]

Thanks in advance!


Looks like the quotient rule.
Reply 3
Original post by Texcoco95
Hi all

As per my screenshot, how do I differentiate from fx to fxx?

Thanks in advance!



Use the quotient rule and keep in mind that x(x2+y2)3/2=32×2x×(x2+y2)3/21\frac{\partial}{\partial x} \left(x^2 + y^2\right)^{3/2} = \frac{3}{2} \times 2x \times (x^2 + y^2)^{3/2 - 1}.
(edited 8 years ago)
Original post by Zacken
Use the quotient rule and keep in mind that x(x2+y2)3/2=32×2x×(x2+y2)3/2\frac{\partial}{\partial x} \left(x^2 + y^2\right)^{3/2} = \frac{3}{2} \times 2x \times (x^2 + y^2)^{3/2}.


how do u square a eqn like this

x<1-\sqrt{x}<1

and how to diffrentiate f(3x) with respect to x ???
Reply 5
Original post by Duke Glacia
how do u square a eqn like this

x<1-\sqrt{x}<1

and how to diffrentiate f(3x) with respect to x ???


Is this related to the thread? :redface:

The first bit isn't an equation...?

ddxf(3x)=3f(3x)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}f(3x) = 3f'(3x).
Original post by Zacken
Use the quotient rule and keep in mind that x(x2+y2)3/2=32×2x×(x2+y2)3/2\frac{\partial}{\partial x} \left(x^2 + y^2\right)^{3/2} = \frac{3}{2} \times 2x \times (x^2 + y^2)^{3/2}.


It should be x(x2+y2)3/2=32×2x×(x2+y2)1/2\frac{\partial}{\partial x} \left(x^2 + y^2\right)^{3/2} = \frac{3}{2} \times 2x \times (x^2 + y^2)^{1/2}
Reply 7
Original post by poorform
It should be x(x2+y2)3/2=32×2x×(x2+y2)1/2\frac{\partial}{\partial x} \left(x^2 + y^2\right)^{3/2} = \frac{3}{2} \times 2x \times (x^2 + y^2)^{1/2}


Whoops! Meant to do 3/213/2 - 1. Thanks!
Original post by Zacken
Whoops! Meant to do 3/213/2 - 1. Thanks!


yeah I always copy over latex and forget to change little things haaaa. Just making it clear to the op who may not have spotted it :smile:

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