1. ## Help Solving Derivatives?

1. y= Ln e^4x - 1 / e^4x + 1

2. F(x) = (tan^2 x - x^2)^3

3. X + Ln x^2 y + 3y^2 = 2x^2 - 1

4. y= Ln(cos3x)

5. F(x) = (x+1)^1/2 - ln [1 + (x+1)^1/2]

And one integral!
(I need to solve this one by substitution) =/

1. |y^3(2y^2-3)dy

2. Originally Posted by mayur03

1. y= Ln e^4x - 1 / e^4x + 1

2. F(x) = (tan^2 x - x^2)^3

3. X + Ln x^2 y + 3y^2 = 2x^2 - 1

4. y= Ln(cos3x)

5. F(x) = (x+1)^1/2 - ln [1 + (x+1)^1/2]

And one integral!
(I need to solve this one by substitution) =/

1. |y^3(2y^2-3)dy
you do not know anything about how to derive functions show some work that's better for you I can solve them but you will not learn try .

$2)f(x)=(\tan ^2 x-x^2)^3$

$f'(x)=3(\tan ^2 x -x^2)^2 (2\tan x (\sec ^2 x) - 2x)$

3. Originally Posted by mayur03

1. y= Ln e^4x - 1 / e^4x + 1

is this supposed to be $\ln\left(\frac{e^{4x}-1}{e^{4x}+1}\right)$ or something else ? ... please use the correct grouping symbols.

2. F(x) = (tan^2 x - x^2)^3

use the chain rule ... twice.

3. X + Ln x^2 y + 3y^2 = 2x^2 - 1

implicit derivative ... is that one term $\ln(x^2y)$ ?

4. y= Ln(cos3x)

derivative of $\ln(u) = \frac{u'}{u}$

5. F(x) = (x+1)^1/2 - ln [1 + (x+1)^1/2]

And one integral!
(I need to solve this one by substitution) =/

1. |y^3(2y^2-3)dy

no substitution is necessary ... distribute the $y^3$ , then integrate.
...