calculus need help badly?

Question

Take the derivative with respect to x: r(x) = sin (x)

take the derivative wth respect to x and smplify completely: fx)=x^1/3(4x^3-sqrt(2x))
please explain with positve exponents.

Answer 1

r ' (x) = cos(x)

f(x) simplify = 4x^1 - { [ (2x)^(1/2) ] * x^(1/3) } = 4x - 2x^(1/2 + 1/3)
= 4x - 2x^ (5/6)
f '(x) = 4 - 2(5/6) x^(5/6 - 1) = 4 - (10/6)x^(-1/6)
x^negative exponent = 1/x^positive exponent
=4 - (10/6) *(1/ (x^(1/6))) = 4 - (10/ (6x^(1/6)))
=4 - (5/ (3x^(1/6)))

Answer 2

r(x) = sin(x) => r'(x) = cos(x)
You simply have to remember the derivative formulas. There is no other way to pass the class.

f(x) = x^(1/3) * (4x^3 - √(2x))
This is a product of two functions g(x) = x^(1/3) and
h(x)=(4x^3 - √(2x)) = (4x^3 - (2x)^(1/2))
and you have to remember how to take the derivative of a function that is a product of two other functions.
f(x) = g(x)g(x)
f'(x) = g(x)h'(x) + g'(x)h(x)
Now
g(x) = x^(1/3) so g'(x) = (1/3)x^(-2/3)
h(x) = (4x^3 - (2x)^(1/2)) => h'(x) = (12x² - (2)(1/2)(2x)^(-1/2))
Now just substitute the expressions for g(x), h(x), g'(x), and h'(x) into the expression for f'(x) above and simplify it.
If you honestly can't do the algebra involved, I would suggest that you really don't have the necessary prerequisites for a Calculus class. This is a pretty basic nuts 'n bolts problem and it only gets deeper from here. Tighten up your algebra and learn the derivative forms.

Doug