Calculus:Chain LInks and derivatives

for the first one, you have to use the chain rule. So the first step is to take the derivative of the outer function, and then do the derivative of the inner function, as follows:
f(x)=(4x-x^2)^100
f’(x)=100(4x-x^2)^99 (derivative of outer function) multiplied by (4–2x) (derivative of inner function).
your teacher may want you to simplify it to:
f’(x)= 200(2-x)(4x-x^2)^99

For ease of typing, dy/dx = y’

1)Find the derivative of this function:
F(x)=(4x-x^2)^100
Answered above

2)Find the first and second derivatives of the function:
y=e^(e^x)
d(e^x)/dy = e^x, so…
lny=e^x
y’/y=e^x
(e^x)(e^e^x)

3)Find dy/dx by implicit differentiation:
sqroot(x+y)=1+(x^2)(y^2)
Square both sides, (x + y) = (1 + 2(x^2)(y^2) + (x^4)(y^4))
1 + y’ = 0 + 2(2x(y^2)+2y’y(x^2)) + (4(x^3)(y^4) + 4(x^4)(y^3)y’)
Simplified to solve for y’
y’ = -(4x(y^2)+4(x^3)(y^4)-1)/(-1+ 4(x^4)(y^3) + 2y(x^2)))

4)Differentiate the function :
f(x)=ln(5th root of x)
y = 1/5 ln(x)
Easy once you realize this, y’ = 1/(5x)

5)Use logarithmic differentiation to find the derivative of the function:
y=sinx^(lnx)
For this one, is it sin(x^lnx) or (sinx)^lnx?

Cheat sheet:
1) 200(2-x)(4x-x^2)^99
2) (e^x)(e^e^x)
3) -(4x(y^2)+4(x^3)(y^4)-1)/(-1+ 4(x^4)(y^3) + 2y(x^2)))
4) 1/(5x)
5) Will answer ASAP

using logarithmic differentiation, first take ln of both sides and simplify, then solve for y’ (you will have a y on the right side). next substitute in the original equation for y.

In this case, if it is y = (sinx)^lnx,
lny = (lnx)*(ln(sinx))
taking the derivative… (this then becomes a implicit differentiation problem)
y^-1*y’ = [(x^-1)(ln(sinx)) + (lnx)((sinx)^-1)*cosx]x’
dy/dx = y[all the x stuff]
next substitute back in the original equation
dy/dx = (sinx)^lnx * [all the x stuff]