derivatives?

Question

Find the derivative of the function.
g(t) = 1 / (t^3 + 3)^4
g'(t) =

Find the derivative of the function.
y = r / sqrt(r^2 + 8)
y' =

Find the derivative of the function. (Remember to enter trigonometric powers such as sin^2x as (sin x)^2.)
f(t) = root3(1+tan t)
f '(t) =

Find the derivative of the function.
y = 7^(1 - x^2)
y' =

Answer 1

The 1'st one is -12t²/(t^3 + 3)^5
Now -you- get your tail in gear and get your homework done. You won't learn a thing if I do it for you ☺

Doug

Answer 2

y= r /sqrt(r^2+8)
y=r (r^2+8)^(-1/2)
let u=r
v=(r^2+8)^(-1/2)
we want udv+vdu
r(-1/2)(r^2+8)^(-3/2)(2r) + (r^2+8)^(-1/2)
-2r^2/(r^2+8)^(3/2) + 1/(r^2+8)^(1/2)

f(t)=(1+tan(t))^(1/3)
f'(t)=(1/3)(1+tan(t))^(-2/3) sec^2(t)
=sec^2(t) / 3 (1+tan(t)^(2/3))

y=7^(1-x^2)
log(y) = (1-x^2)log(7)
(1/y)dy/dx = log(7) (-2x)
dy/dx=-2xy log(7)
=-2xlog(7)7^(1-x^2)