A Level Maths Differentiation Question

https://www.quora.com/profile/Bravewarrior/p-162563693
Here is the question. I've tried so many ways of getting to what they need, but for some reason I'm not able to. Can someone please help me with this question?
Thanks 🙂

Reply 1

mqb2766

21

Original post by pigeonwarrior

https://www.quora.com/profile/Bravewarrior/p-162563693
Here is the question. I've tried so many ways of getting to what they need, but for some reason I'm not able to. Can someone please help me with this question?
Thanks 🙂

maybe upload what youve tried? You could do the denominator as a power of ^(-1) or use the quotient rule.

Reply 2

davros

16

Original post by pigeonwarrior

https://www.quora.com/profile/Bravewarrior/p-162563693
Here is the question. I've tried so many ways of getting to what they need, but for some reason I'm not able to. Can someone please help me with this question?
Thanks 🙂

Have you come across logarithmic differentiation? It won't necessarily be faster, but all roads should lead to the same result, so if you can do it via the quotient rule (f = u/v) then you should certainly be able to reproduce the answer with the product rule and chain rule (f = uv^-1).

(So you basically have f = uvw/z where u = 7, v = x, w = e^x and z = the horrible bit on the bottom! Then taking logs of both sides gives you ln f = ln u + ln v + ln w - ln z. Now differentiate w,r,t.x to get (1/f)(df/dx) = (1/u)du/dx + (1/v)(dv/dx) + (1/w)(dw/dx) - (1/z)(dz/dx). Sort out the derivatives on the RHS and put over a common denominator, then multiply both sides by f to get df/dx on the left. Should work out - I've just tried it in rough and it looks OK 🙂 )

Reply 3

davros

16

Original post by pigeonwarrior

https://www.quora.com/profile/Bravewarrior/p-162563693
Here is the question. I've tried so many ways of getting to what they need, but for some reason I'm not able to. Can someone please help me with this question?
Thanks 🙂

Have you come across logarithmic differentiation? It won't necessarily be faster, but all roads should lead to the same result, so if you can do it via the quotient rule (f = u/v) then you should certainly be able to reproduce the answer with the product rule and chain rule (f = uv^-1).

(So you basically have f = uvw/z where u = 7, v = x, w = e^x and z = the horrible bit on the bottom! Then taking logs of both sides gives you ln f = ln u + ln v + ln w - ln z. Now differentiate w.r.t.x to get (1/f)(df/dx) = (1/u)du/dx + (1/v)(dv/dx) + (1/w)(dw/dx) - (1/z)(dz/dx). Sort out the derivatives on the RHS and put over a common denominator, then multiply both sides by f to get df/dx on the left. Should work out - I've just tried it in rough and it looks OK 🙂 )

Reply 4

pigeonwarrior

OP

11

Thank youuu so much everyone! After a very very long time I have finally made it to what the question wanted! Thank you for all the tips! 🙂🙂🙂

(edited 11 months ago)

A Level Maths Differentiation Question

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