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Core 3 Differentiation

Hi,
I know this is probably an easy question but how would you go about differentiating 4ln2x? Thanks
Original post by student1856
Hi,
I know this is probably an easy question but how would you go about differentiating 4ln2x? Thanks


Use the Chain rule.
Reply 2
Original post by zetamcfc
Use the Chain rule.


What part of it would I set as U? Would it be y=4U with U equalling ln2x?
Original post by student1856
What part of it would I set as U? Would it be y=4U with U equalling ln2x?


u=2x.
Reply 4
Original post by student1856
Hi,
I know this is probably an easy question but how would you go about differentiating 4ln2x? Thanks


y=ln(ax) y=ln(ax)
dydx=aax=1x \dfrac{dy}{dx} = \dfrac{a}{ax} =\dfrac{1}{x}
(edited 7 years ago)
4 is a constant,
dy/dx(Inx) would be 1/x


as for ln2x:

dy/dx(In2x) = dy/dx(Inu) = 2/u = 2/2x = 1/x

so, dy/dx = constant/x.
(edited 7 years ago)
Reply 6
Original post by zetamcfc
u=2x.


Right, so I do that and I get 2 multiplied by 4/x which gives me 8/x. But the answer is 4/x.
Original post by student1856
Right, so I do that and I get 2 multiplied by 4/x which gives me 8/x. But the answer is 4/x.

read above ^
Original post by student1856
Right, so I do that and I get 2 multiplied by 4/x which gives me 8/x. But the answer is 4/x.


y=4lnu implies dy/dx = 4/u * du/dx
Reply 9
Original post by XOR_
4 is a constant,
dy/dx(Inx) would be 1/x


as for ln2x:

dy/dx(In2x) = dy/dx(Inu) = 2/u = 2/2x = 1/x

so, dy/dx = constant/x.



So, why is 4 a constant here but when you calculate 4lnx, 4 is not a constant? Why do you include the 4 in the diffeentiationof 4lnx but not in 4ln2x? Thank you
Original post by student1856
So, why is 4 a constant here but when you calculate 4lnx, 4 is not a constant? Why do you include the 4 in the diffeentiationof 4lnx but not in 4ln2x? Thank you

dy/dx(4lnx) = 4/x

dy/dx(4ln2x) = 4/x

for dy/dx(ln2x),
when u = 2x, you get dy/dx(lnu) = 2/2x = 1/x.
(edited 7 years ago)
Reply 11
Original post by student1856
So, why is 4 a constant here but when you calculate 4lnx, 4 is not a constant? Why do you include the 4 in the diffeentiationof 4lnx but not in 4ln2x? Thank you


Would it help if we did this:

4ln(2x)=4(ln2+lnx)=4ln2+4lnx4\ln(2x) = 4(\ln 2 + \ln x) = 4\ln 2 + 4\ln x.

So: ddx(4ln2+4lnx)=0+4ddx(lnx)\frac{d}{dx}(4\ln 2 + 4\ln x) = 0 + 4\frac{d}{dx}(\ln x)
Original post by XOR_
dy/dx(4lnx) = 4/x

dy/dx(4ln2x) = 4/x

for dy/dx(ln2x),
when u = 2x, you get dy/dx(lnu) = 2/2x = 1/x.


Ahh, I get it now. Thank you very much
Original post by Zacken
Would it help if we did this:

4ln(2x)=4(ln2+lnx)=4ln2+4lnx4\ln(2x) = 4(\ln 2 + \ln x) = 4\ln 2 + 4\ln x.

So: ddx(4ln2+4lnx)=0+4ddx(lnx)\frac{d}{dx}(4\ln 2 + 4\ln x) = 0 + 4\frac{d}{dx}(\ln x)


Yes it does, thanks Zacken
Reply 14
Original post by student1856
Yes it does, thanks Zacken


:smile:

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