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# y = 5e^3x watch

1. Find the gradient of the curve y=5e^3x at the point for which x= (ln a)

Could someone please explain why the answer is 15a^3 ????
2. What do you get when you differentiate that? Once you get dy/dx, that's you gradient formula, you just plug in the x value and you will get 15a^3
3. (Original post by FoOtYdUdE)
Find the gradient of the curve y=5e^3x at the point for which x= (ln a)

Could someone please explain why the answer is 15a^3 ????
What did you get when you differentiated it?
Edit: too slooooooooow
4. The differential wrt x is 15e^3x.

This gives the rate of change of y wrt x, i.e. the gradient, at some x.

At x=lna, 3x=3lna=ln(a^3)

Thus the gradient is 15*e^[ln(a^3)]=15a^3
5. remember that e^kx differentiated is ke^kx
6. 15e^3x, then put sub x with lna
7. (Original post by Mr.Singh)
15e^3x, then put sub x with lna
yeah, i get the 15e^3x part....but u know when you subsititue it...i don't see why the 3 remains as a power and the a comes down.
8. (Original post by FoOtYdUdE)
yeah, i get the 15e^3x part....but u know when you subsititue it...i don't see why the 3 remains as a power and the a comes down.
k ln x = ln(x^k)
Can you do it from there?
9. ok, so 15e^(3 x ln a)

i get stuck after writing that.
10. when u have the power=3lna
the 3 goes to the front as a power of a. the e and ln are like them inversey thingies so they cancle leaving 15a^3 cos the 3 is still the power of a
11. (Original post by FoOtYdUdE)
ok, so 15e^(3 x ln a)

i get stuck after writing that.

12. OMG guys, i love you!!!!!!! Thank you. Sorry for being so dumb.
13. footy dude. look at the rules of e and ln on this website. its good for revision http://www.mathsrevision.net/alevel/pure/
14. (Original post by Naffy)
you guys are making it too confusing.

e and ln cancel out, thats all there is too it (ignore the above) no offense to 2710, but you're overcomplicated a 'relatively' simple C3 question
What are you talking about? In my post I showed that e and ln cancel out. I dont think theres anything hard about it...
15. (Original post by Naffy)
you guys are making it too confusing.

e and ln cancel out, thats all there is too it (ignore the above) no offense to 2710, but you're all overcomplicating a 'relatively' simple C3 question

ln a x e = a
i think you'll find that's not true.

For example:

ln(3) * e = 2.98633782
ln(6) * e = 4.87050721

However,

16. y=5e^3x
dy/dx=15e^3x
when x=ln(a)
dy/dx=15e^3ln(a)--->e^ln cancel each other out so that a goes down
so you get: dy/dx=15a^3
17. (Original post by zdo0o)
i think you'll find that's not true.
I noticed my mistake, thats why I deleted straight after. I misunderstood that you guys were trying to explain the change of power log law.

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