Use suitable derivative rules to find y' if:?

Question

(a) y = -7x*(-5/3)

(b) y = -(1/x^2) - sqroot x + (1/2)

(c) y = (x^3 + 4)(-x^6 + x^4) (Use Product rule?)

(d) y = (sqroot x + (1/2))(x^3 + x^(1/3) (Use product rule?)

(e) y = 2x^3 + 4 / x + 7

(f) y = x^(2/3) + 3 / x^(1/5) + x

I don't even know which rules to use where, or what the product rule is.

Answer 1

your main rule for finding derivatives is:
y = x^n
y' = nx^(n-1)

translated, to find the derivative of x to some power n, you multiply the coefficient of x by that power (if its just x, the coefficient is 1, if it's say, 4x, then you multiply n * 4) then subtract 1 from the exponent, thus n-1

for your first example a)
y = -7x^(-5/3) in this case, n= -5/3 and the coefficient of x is -7. So, you'd multiply -7 by -5/3 and that becomes x's new coefficient. Then you subtract 1 from -5/3 for the new exponent. And you've got the derivative.

When it comes to square roots and fractions, just bring them all into the same format and you can use this rule. Example: v( ) denotes square root
v(x) is the same as x^(1/2)
4/x is the same as 4(x^-1)
1/v(x) is the same as x^(-1/2)

if you have a square root, its the same as multiplying the inside of the square root by 1/2
if anything is in the denominator of the fraction, you can bring it to the top and change the power to a negative. Then solve by your first rule.

The product rule is, for two functions g(x) and f(x)
f(x)*g'(x) + g(x)*f'(x)

for example, in c, the two functions your are multiplying are separated by parenthesis. So, you can call one f(x) and one g(x). Let's say its like this:
f(x) = x^3 + 4
g(x) = -x^6 + x^4

So first you'll have to find the derivative of both (g'(x) and f'(x)) and then plug that in to the product rule equation.

And don't forget, the derivative of a constant (any number without a variable like x) turns to 0 in the derivative.
y = x^2 + 4
y' = 2x^1 (or just 2x)

Also, the derivative of x without any other exponents is just 1
y = x
y' = 1

Answer 2

(a) y = -7x*(-5/3)
y'=35/3x^-8/3

(b) y = -(1/x^2) - sqroot x + (1/2)
y'=2x^(-1)-1/(2sqrt(x))+0

(c) y = (x^3 + 4)(-x^6 + x^4) (Use Product rule?) yes
y'=3x^2((-x^6 + x^4)+(-6x^5+4x^3)( (x^3 + 4)

Very easy. Continue to finish them.

Answer 3

some more.
e) y=2x^2+4/x+7
y'=2(2)x-4/x^2
=4x-4/x^2
Note:
derivative if 1/x= x^(-1) = (-1)x^(-2)=-1/x^2