What are the derivatives of these two functions?

Question

From what I understand, the derivative formula is: nx^(n-1)
Example,
f(x) = 4x^2
f'(x) = 4x^2
= 4 * 2 * x^(2-1)
= 8x

1) y = a * e^v + (b/v) + (c/(v^2)), find y'.
Here's what I've got so far:

y' = 1 * ve^(v-1) + (1/1) + (1/(2v))
= ve^(v-1) + 1 + (1/2)v

Where do I go from here (assuming that is right)?

2) z = A/(y^10) + B * e^y, find z'.
Again:

z' = 1/(10y^9) + 1 * ye^(y-1)

When you have a variable to the power of 1, that means it can be canceled, right? f(x) = B, f'(x) = 1*B^(1-1) = 1?

Oh, one more question: Is the cubic root of x squared the same as x^(3/2) or is it x^(2/3)?

Answer 1

1) I'm assuming you are taking the derivative with respect to v.
The derivative of e^v is (simply) e^v. You do not subtract 1 from the exponent of the exponential (e).
For the other two terms, you must rewrite it so that the variable is in the numerator (and then you can apply your formula)
b/v = b v^(-1)
c/v^2 = c v^(-2)
y' = a*e^v + b*(-1)*v^(-1-1) + c*(-2)*v^(-2-1)
y' = a*e^v - b*v^(-2) - 2*c*v^(-3)
You may wish to rewrite it as
y' = a*e^v - b/v^2 - 2c/v^3

Now try #2

3) f(x) = B
B is a constant, and the derivative of a constant is always zero.
f'(x) = 0

4) The cubic root of x squared is x^(2/3)
x^(3/2) would be the square root of x cubed.

Answer 2

In your first equation, it looks like the variable is supposed to be v.

y=ae^v + b/v + c/v^2

The derivative of e^x is e^x, so ae^v would still be ae*v. b/v and c/v^2 can be re-written with negative exponents: bv^(-1) and cv^(-2). So then you can apply the shortcut on those:

bv^(-1) would be (-1)*b*v^(-1-1) which simplifies to -b*v^(-2) or, moving the negative exponents back to the denominator: -b/(v^2).

cv^(-2) would be -2*c*v^(-2-1) or -2c/(v^3)

So y'= ae^v -b/v^2 -2c/v^3

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Now for number 2 - it looks like y is your variable. Again, Be^y stays Be^y. Rewrite A/(y^10) as Ay^(-10) and do it the same way: -10*A*y^(-10-1) = -10Ay^(-11) or -10A/(y^11).

So the final answer would be y' = -10A/y^11 + Be^y

~~~~~~~~~~~~~~~~~~~~~~

As for your questions:

Yes, a variable to the first power always differentiates to 1. If y=x then y' = 1. If y=2x then y' = 2*1 = 2

The cube-root of x^2 is x^(2/3)

Remember how (x^2)^3 = x^6? Because when you have a power raised to a power you multiply the exponents. Well the cube root of x is x^(1/3). Then you square it (x^1/3)^2. Multiply the powers 1/3 * 2 = 2/3.

Hope that helps some.

Answer 3

Oh, one more question: Is the cubic root of x squared the same as x^(3/2) or is it x^(2/3)?
x^(2/3)

When you have a variable to the power of 1, that means it can be canceled, right? f(x) = B, f'(x) = 1*B^(1-1) = 1?
Right man but B is a constant
Also when is Ax for example, the constant does not cancel (Ax)' = A

Answer 4

the rule d(x^n) = n x^(n-1) applies only to variables as base (x is variable, n is constant); and not to e^x (e is constant, x is variable)

we know d(e^x) = e^x (unchanged)

1/x^n can be written as x^(-n)
hence d(1/x^n) = -n x^ (-n-1) = -n x^(-(x+1)) = -n / x^(n+1)


so y=a * e^v + (b/v) + (c/(v^2))

d(y) = a * e^v - (b/v^2) -2c/v^3

similarly

z = A/(y^10) + B * e^y
d(z) = -10A/y^11 + B*e^y

Answer 5

It's been a while since I've done derivatives, but I think that in your examples, a b and c are constants, not variables, so they would remain. For example:

f(x) = ax^2
f'(x) = 2ax

That's your first mistake.

Answer 6

Note that derivative of e^x = e^x and 1/x^2 is equal to x^-2.
Therefore your solution for 1-) a.e^v - b/v^2 -2c/v^3

And for 2-) -10A/y^11 + Be^y

Answer 7

'fraid not:
y=e^x; y'=e^x
y=a/(x^n); y'=-n*a/(x^n+1)

1) y'=a*e^v-b/(v^2)-2*c/(v^-3)

2) z'=-10*A/(y^11)+B*e^z

3) y=B=b*x^0; y'=0*B=0

4) y=Bx; y'=B