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C3 Differentiation Problem

I'm stuck on question 7 part iii of this past paper
https://7c95ed1f832973eac0c6e1a52b08181fd38e50f6.googledrive.com/host/0B1ZiqBksUHNYcWYycEk2VmM2b00/January%202011%20QP%20-%20C3%20OCR.pdf

f(x) = ln x
g(x) = x^2 + 8

Find the gradient of the curve gf(x) where x = e^3

I can get to (ln x)^2 + 8 but every time I try to use the chain or product rule i always get it wrong
Reply 1
Avatar for Zacken
Zacken
22
Original post by JW22
I'm stuck on question 7 part iii of this past paper
https://7c95ed1f832973eac0c6e1a52b08181fd38e50f6.googledrive.com/host/0B1ZiqBksUHNYcWYycEk2VmM2b00/January%202011%20QP%20-%20C3%20OCR.pdf

f(x) = ln x
g(x) = x^2 + 8

Find the gradient of the curve gf(x) where x = e^3

I can get to (ln x)^2 + 8 but every time I try to use the chain or product rule i always get it wrong


ddx((ln⁡x)2+8)=2(ln⁡x)2−1×ddx(ln⁡x)\displaystyle \frac{d}{dx} ((\ln x)^2 + 8) = 2 (\ln x)^{2-1} \times \frac{d}{dx} (\ln x)
Original post by JW22
I'm stuck on question 7 part iii of this past paper
https://7c95ed1f832973eac0c6e1a52b08181fd38e50f6.googledrive.com/host/0B1ZiqBksUHNYcWYycEk2VmM2b00/January%202011%20QP%20-%20C3%20OCR.pdf

f(x) = ln x
g(x) = x^2 + 8

Find the gradient of the curve gf(x) where x = e^3

I can get to (ln x)^2 + 8 but every time I try to use the chain or product rule i always get it wrong

Wouldn't it just be 2*(lnx)*(1/x), and if you sub e^3 in you would get (2*(3lne))/e leading to 6/e^3 because lne = 1. I just did this in my head quickly I may have made a mistake
(edited 8 years ago)
Reply 3
Avatar for JW22
JW22
OP
10
Original post by Zacken
ddx((ln⁡x)2+8)=2(ln⁡x)2−1×ddx(ln⁡x)\displaystyle \frac{d}{dx} ((\ln x)^2 + 8) = 2 (\ln x)^{2-1} \times \frac{d}{dx} (\ln x)


Thanks
Reply 4
Avatar for Zacken
Zacken
22
[QUOTE=OrionMusicNet;63636081[non-beneficial stuff]


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