QUESTION about "derivatives"?

Question

An easy question about derivatives, but this is not homework.

d/dx [πx³ - 2(√x)¹³ + 7e - π²] = ?

^_^

Answer 1

ANSWER!!!

3πx²- 13√x¹¹

Pretty simple ah!!!
We did'nt include the derivative of 7e and -π² because its just a ZERO!! no nid for that!!! Why zero?? Because its a constant number and doesn't have a variable!!

How to get the answer!!
Derivative = nx^n-1
Substitute
πx³
3πx²

For
2(√x)¹³
13√x¹¹
too long to explain i know that you know it your a genius hihihi ^_^!!!

Answer 2

d/dx [πx³ - 2(√x)¹³ + 7e - π²] =

3πx^2-(13)*x^(11/2)

remember when u differentiate a constant you get zero. you can write √x as x^(1/2), so (√x)¹³=x^(13/2).

Answer 3

d/dx [πx³ - 2(√x)¹³ + 7e - π²] =3 πx^2 - 13(√x)^{11}

Answer 4

d/dx [πx³ - 2(√x)¹³ + 7e - π²] = 3πx² - 13(√x)¹¹

Answer 5

Remember that:

*) Derivatives are linear, so d( a*f(x) + b*g(x) )/dx = a*f'(x) + b*g'(x). Among other things, this means that you can take derivatives of each of the individual terms and add them together.

*) Derivatives of functions of the form x^n are given by n*x^(n-1)

*) Taking the square root of x (denoted sqrt(x)) is the same as raising x to the 1/2, so sqrt(x)^13 is ( x^(1/2) )^13 = x^(13/2).

*) Derivatives of constants are 0.

Therefore, your derivative is:

3*TT*x^2 + 2*(13/2)*x^(11/2)

which you can simplify to get:

3*TT*x^2 + 13*sqrt(x)^11