# Thread: Derivatives of Logarithmic Functions

1. ## Derivatives of Logarithmic Functions

Find the derivative:
1) y=x^(x^3)
2) y=(sin x)^(x^2)

Find dy/dx implicitly:
1) 4xy+ln(x^2*y)=7

Thanks a lot!

2. Originally Posted by cupcakelova87
Find the derivative:
1) y=x^(x^3)
2) y=(sin x)^(x^2)
I suppose you want to find these using logarithmic differentiation, here goes:

1) y=x^(x^3)
=> ln(y) = ln[x^(x^3)] ..................log both sides
=> ln(y) = (x^3)ln(x) ....................use a law of logs to bring the power down, log(x^n) = nlog(x)
=> y'/y = 3(x^2)ln(x) + x^2 ..........differentiate both sides with respect to x, use product rule for the right hand side
=> y'/y = (x^2)(3ln(x) + 1) ...........simplify a bit
=> y' = y[(x^2)(3ln(x) + 1)] ..........solve for y'
=> y' = [x^(x^3)][(x^2)(3ln(x) + 1)]....write y' in terms of x

2) y=(sin x)^(x^2) ......................same procedure
=> ln(y) = ln[(sin x)^(x^2)]
=> ln(y) = (x^2)ln(sin(x))
=> y'/y = 2xln(sin(x)) + (x^2)cos(x)/sin(x)
=> y' = y[2xln(sin(x)) + (x^2)cos(x)/sin(x)]
=> y' = [(sin x)^(x^2)][2xln(sin(x)) + (x^2)cos(x)/sin(x)]

3. Originally Posted by cupcakelova87
Find dy/dx implicitly:
1) 4xy+ln(x^2*y)=7
1) 4xy + ln(x^2*y) = 7
=> 4y + 4x y' + [1/(x^2 *y)]*(2xy + x^2 y') = 0
=> 4x y' + [x^2/(x^2 *y)]y' = -4y - 2xy/(x^2 *y)
=> y' {4x + [x^2/(x^2 *y)]} = -4y - 2xy/(x^2 *y)
=> y' = [-4y - 2xy/(x^2 *y)]/{4x + [x^2/(x^2 *y)]}
=> dy/dx = [-4y - 2xy/(x^2 *y)]/{4x + [x^2/(x^2 *y)]}

and you can simplify that a bit if you want

4. Hello, cupcakelova87!

Find dy/dx implicitly: .4xy + ln(x²y) .= .7
I would simplify that log first . . .

4xy + ln(x²) + ln(y) .= .7 . . . .4xy + 2·ln(x) + ln(y) .= .7

Differentiate: .4x·y' + 4y + 2/x + y'/y .= .0

Then we have: .4x·y' + y'/y .= .-4y - 2/x

Factor: .(4x + 1/y)·y' .= .-2(2y + 1/x)

. . . . . . . . . . . 4xy + 1 . . . . . -2(2xy + 1)
And we have: . --------- · y' .= .--------------
. . . . . . . . . . . . . .y . . . . . . . . . .x

. . . . . . . . . . . . . .-2y(2xy + 1)
Therefore: .y' . = . ---------------
. . . . . . - . . . . . . . x(4xy + 1)

I messed up my signs.
They're correct now (I hope).

5. Originally Posted by Soroban
Hello, cupcakelova87!

I would simplify that log first . . .

4xy + ln(x²) + ln(y) .= .7 . . . .4xy + 2·ln(x) + ln(y) .= .7

Differentiate: .4x·y' + 4y + 2/x + y'/y .= .0

Then we have: .4x·y' + y'/y .= .-4y - 2/x

Factor: .(4x + 1/y)·y' .= .2(1/x - 2y)

. . . . . . . . . . . 4xy + 1 . . . . . 2(1 - 2xy)
And we have: . --------- · y' .= .-----------
. . . . . . . . . . . . . .y . . . . . . . . . .x

. . . . . . . . . . . . . .2y(1 - 2xy)
Therefore: .y' . = . --------------
. . . . . . - . - . . . . .x(4xy + 1)

yeah, usually i'd do that too, but i'm all scatter-brain today for some reason. thanks for looking out

6. Originally Posted by Soroban
Hello, cupcakelova87!
Then we have: .4x·y' + y'/y .= .-4y - 2/x

Factor: .(4x + 1/y)·y' .= .2(1/x - 2y)
Oops... you forgot a minus sign there...