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Calculus Help!

Hi guys, I really need some help with my calculus.
Iv'e searched on the internet to help with my understanding but I can't get my head around it!

The two questions I have are attached as a picture.

I would love some help, thanks in advance!

FullSizeRender.jpg
follow the rule
d(axn)dx=anxn−1\frac{d(ax^{n})}{dx}=anx^{n-1}
for example
d(5x7)dx=35x6\frac{d(5x^{7})}{dx}=35x^{6}
or
d(5x75)dx=7x25\frac{d(5x^{\frac{7}{5}})}{dx}=7x^{\frac{2}{5}}
(edited 7 years ago)
Reply 2
Avatar for TomNugentt
TomNugentt
OP
Original post by depymak
follow the rule
d(axn)dx=anxn−1\frac{d(ax^{n})}{dx}=anx^{n-1}
for example
d(5x7)dx=35x6\frac{d(5x^{7})}{dx}=35x^{6}
or
d(5x75)dx=7x25\frac{d(5x^{\frac{7}{5}})}{dx}=7x^{\frac{2}{5}}


Would you be able to give an example similar to one of those questions to help my understanding?
Reply 3
Avatar for B_9710
B_9710
18
d(x−1/2)dx=−12x−3/2 \displaystyle \frac{d(x^{-1/2})}{dx} = -\frac{1}{2} x^{-3/2} .
Reply 4
Avatar for B_9710
B_9710
18
Original post by depymak
follow the rule
d(axn)dx=anxn−1\frac{d(ax^{n})}{dx}=anx^{n-1}
for example
d(5x7)dx=35x6\frac{d(5x^{7})}{dx}=35x^{6}
or
d(5x75)dx=7x25\frac{d(5x^{\frac{7}{5}})}{dx}=7x^{\frac{2}{5}}


Last one, there's a slight mistake.
Original post by B_9710
Last one, there's a slight mistake.

Could you please explain?
Original post by B_9710
Last one, there's a slight mistake.


it looks OK to me :holmes:
Reply 7
Avatar for B_9710
B_9710
18
Original post by depymak
Could you please explain?


It's ok, my mistake. For some reason I though the exponent was 7/2 rather than 7/5 so I'm my head I was wrong. Never mind. :colondollar:
Reply 8
Avatar for B_9710
B_9710
18
Original post by the bear
it looks OK to me :holmes:


Yeah my mistake buddy.
You know where you struggle with KS2 SATS subtraction for a minute or 2.

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