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differentiating forms of a^x

hey guys I'm having trouble understanding how to differentiate a constant to the power of a variable, for example

y=5^x
I know
dy/dx = 5^x ln(5)

and this is the same for any integer, but I don't understand the mechanism and get stuck when it's something of the form

5x^x or 5^x^2
see I don't get how this follows :

y = x^5
ln Y = X ln (5)

differntiate both sides,

= 1/y = (x/5) + ln(5)

and then i'm confused.
Reply 2
Avatar for steve2005
steve2005
17
Original post by kingkongjaffa
hey guys I'm having trouble understanding how to differentiate a constant to the power of a variable, for example

y=5^x
I know
dy/dx = 5^x ln(5)

and this is the same for any integer, but I don't understand the mechanism and get stuck when it's something of the form

5x^x or 5^x^2


Take logs and then differentiate.
I meant 5^x not x^5 which of course is 5x^4...
Original post by steve2005
Take logs and then differentiate.


can you show me I get stuck after taking logs?
Are you taking logs and then using implicit differentiation?

y=5x2y=5^{x^2}

ln⁡y=x2ln⁡5\ln y = x^2 \ln 5

and so on?
Original post by Mr M
Are you taking logs and then using implicit differentiation?

y=5x2y=5^{x^2}

ln⁡y=x2ln⁡5\ln y = x^2 \ln 5

and so on?


yeah i get stuck there the logs isn't a problem
Reply 7
Avatar for Blutooth
Blutooth
16
Original post by kingkongjaffa
can you show me I get stuck after taking logs?


Here's a quicker way to do it.
1)Note that ax=(elna)x=elna⋅xa^x=(e^{lna})^x=e^{lna\cdot x}
2)Remember that ebx⇒dydx=bebxe^{bx} \Rightarrow \frac{dy}{dx}=be^{bx}
3) Letting b=ln(a)b=ln(a) yields dydx=ln(a)⋅elna⋅x=ln(a)⋅ax \frac{dy}{dx}=ln(a) \cdot e^{lna\cdot x}=ln(a)\cdot a^x
(edited 12 years ago)
Original post by kingkongjaffa
can you show me I get stuck after taking logs?


As a general rule:

y = a^x

ln(y) = ln(a^x)

ln(y) = x ln(a)

Now differentiating - the differential of ln(x) is 1/x, so the differential of ln(y) is going to be 1/y times dy/dx:

1/y * dy/dx = ln(a) <- differentiated the RHS by treating ln(a) as a constant, just as 3x goes to 3 when you diff.

dy/dx = y ln(a)

but y = a^x

so dy/dx = a^x ln(a)
Original post by kingkongjaffa
yeah i get stuck there the logs isn't a problem


ln⁡y=x2ln⁡5\ln y =x^2 \ln 5

1ydydx=(2ln⁡5)x\frac{1}{y} \frac{dy}{dx} = (2\ln 5) x

You know what y = ...?
ok i realised what i was doing wrong and understand how to do it in the form a^x without looking at the rule....
y=5^x
lny=ln(5^x)
lny=xln5

differentiate both sides, remembering ln5 is constant
Reply 12
Avatar for kingfisher
kingfisher
How will one apply chain rule when its related to a fraction in a square root?

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