Question

any calculus guys out there? how do you find the first derivative of z^2+1/3z

Answer 1

derivative of z^2 is 2x.
derivative of 1/z is -1/z^2

You should be able to do the rest.

Answer 2

\[\huge \frac{\partial (z^2+\frac{1}{3}z)}{\partial z} = 2z+\frac{1}{3}\]

Answer 3

d/dz(z^2) + d/dz(1/3z)
2z-1/3z^2

Answer 4

f' = 2z + (1/3) * - z^-2
= 2z - 1/3z^2

Answer 5

\[\huge \frac{\partial (z^2+\frac{1}{3}z^{-1})}{\partial z} = 2z-\frac{1}{3}z^2\]

Answer 6

For some number n, \[\huge \frac{\partial (z^n)}{\partial z} = nz^{n-1}\]

Answer 7

i guess the question is, is this a local toy store sells whistles in packages of 3,6,or 12. if you need 30 whistles for a party, in how many diffrent combinations can you buy them?
\[z^2+\frac{1}{3}z\] or
\[z^2+\frac{1}{3z}\]

Answer 8

cyter, the symbols you are using should be reserved for partial derivatives. A plain script d is the application here.

Answer 9

in the back of my book it says that the answer is (z^2-1)/3z^2. i just don't understand how to get that answer but i know that i am supposed to use the power rule

Answer 10

that is because ( i am just guessing) that your problem is
\[f(z)=\frac{z^2+1}{3z}\]

Answer 11

@satellite73 - what syntax tool do you use to write equations in that format? Much cleaner to see than flat text writing.

Answer 12

no one actually answered that question. the answers here were to
\[\frac{d}{dz}[z^2+\frac{1}{3}z]\] and
\[\frac{d}{dz}[z^2+\frac{1}{3z}]\]

Answer 13

@gt latex

Answer 14

Use the quotient rule

Answer 15

ok i solved it using the quotient rule, but since im guessing my calc teacher will have me solve a problem like that using only the power rule on the test. so can anyone solve it using the power rule and getting the correct answer that i got in my book, and show me how to do it?

Answer 16

ok it would really help if we knew exactly what the problem was

Answer 17

to find the derivative of that function, you must use both the power rule and the quotient rule

Answer 18

i repeat, it would really be helpful to know what "that function" actually is. maybe we can use nothing but the "power rule"... maybe