Find dy/dx for the following functions. you need not simplify your answer.?

Question

1. y= sin^3 (2x + e^x)

2. y= (x^2 - 7x + 1)/(5x^3 - x)

3. y= x^(cosx)

plz help me for my calc class

Answer 1

1. dy/dx=3sin(2x+e^x)^2(2+e^x)

2. ((2x-7)(5x^3-x)-(15x^2-1)(x^2-7x+1))/(5x...

3. not sure about that one.

make it a good day

Answer 2

1)  y'=3sin^2(cos)(2x+e^x)+sin^3(2+e^x)

2)  y'=(2x-7)(5x^3-x)-(x^2-7x+1)(15x^2-1)/25...

    y'=10x^4-2x^2-35x^3+7x-(15x^4-x^2-105x^3...

y'=-5x^4+14x^2+70x^3-1/15x^6-10x^4+x^2

Answer 3

It looks like yahoo doesn't like long equations, so I broke some of mine up, don't miss the end of the equations!

1. Use the chain rule. The outermost equation is x^3, which contains sin(x), which contains 2x+e^x

y'=3(sin(2x+e^x))^2

*cos(2x+e^x)*(2+e^x)

2. Use the quotient rule

y'= ((5x^3 - x)*d/dx[x^2 - 7x + 1]-(x^2 - 7x + 1)*d/dx[5x^3 - x])

/((5x^3 - x)^2)

y'= ((5x^3 - x)*(2x - 7)-(x^2 - 7x + 1)*(15x^2 - 1))

/((5x^3 - x)^2)

If you wrote the third one correctly, it's really wacky.  First I had to look up what d/dx of x^x was, that's the first source, please check it out to be sure you follow.

y=x^cosx

take the natural log of both sides, then use log rules to simplify

ln(y)=ln(x^cosx)=cosx*ln(x)

take the derivative of each side

d/dx[ln(y)]=d/dx[cosx*ln(x)]

use the chain rule on the left and the product rule on the right

(1/y)*(dy/dx)=

(cosx*d/dx[ln(x)]+ ln(x)*d/dx[cosx])

(1/y)*(dy/dx)=(cosx*(1/x)+ ln(x)*(sin^-(x))

recall that dy/dx=y'.  Multiply both sides by y and recall that y=x^cosx

y'=x^cosx(cosx/x

+ln(x)sin^-(x))