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Calculus1 21 Online
OpenStudy (anonymous):

if f(z)=az+b and g(z)=cz+d, what are f'(z) and g'(z)

OpenStudy (loser66):

z is variable, a, b are constants

OpenStudy (loser66):

so that, take derivative respect to variable only

OpenStudy (anonymous):

sorry all the z's are accually x's

OpenStudy (loser66):

doesn't matter, the same

OpenStudy (anonymous):

I haven't done a derivative in 16 years I'm lost

OpenStudy (loser66):

want to re study or just the answer?

OpenStudy (anonymous):

i need both

OpenStudy (loser66):

hahaha... ok be patient, I will break it to baby steps. Ready?

OpenStudy (anonymous):

go for it

OpenStudy (loser66):

(1)'=0 (2)'=0 (3)'=0......... so that if you have to take derivative of a number (so-called constant), the result =0 , got me? now test, as above, I said that a, b is number, (b)' =?

OpenStudy (loser66):

knock, knock!! hard to understand or I gave you a preschool step? hahahaha ....

OpenStudy (anonymous):

wow I think I lost it! I'm amazed I can't remember any of this

OpenStudy (loser66):

no need to remember any thing, I started a brand new lecture. hihihihii... read my comment and answer me (b)'=?

OpenStudy (anonymous):

b'=0

OpenStudy (loser66):

Question: why do you have to take the course when you forgot all knowledge? This is the calculus1's stuff. The basis step of calculus.

OpenStudy (loser66):

yup!

OpenStudy (loser66):

now, variable: if x is variable, then (x)'= 1 if t is a variable, then (t)'=1 how about if y is a variable?

OpenStudy (anonymous):

y'=1

OpenStudy (loser66):

yup again

OpenStudy (loser66):

now, if number* variable, and we have to take derivative of THEM, for example, 2x, it is 2*x and we just take derivative of the variable only, (2x)'=2 because x'=1 and 2*1 =2 right? Now your turn (5y)'=?

OpenStudy (anonymous):

y'=5

OpenStudy (loser66):

you miss 5 in the front of y', redo

OpenStudy (anonymous):

so it's the slope

OpenStudy (anonymous):

y'=1

OpenStudy (loser66):

5y'=?

OpenStudy (anonymous):

=5

OpenStudy (loser66):

good, so, if a is number, x is variable and (a*x)' =?

OpenStudy (anonymous):

=a

OpenStudy (loser66):

bingo

OpenStudy (loser66):

one more thing, sum of derivative = derivative of sum. That means (ax + b)' = (ax)' +b' got me? so, (ax+b)'= ?

OpenStudy (anonymous):

a+b

OpenStudy (loser66):

b is a number, and derivative of number is = ??

OpenStudy (anonymous):

0

OpenStudy (loser66):

so, redo, (ax+b)'=

OpenStudy (anonymous):

a

OpenStudy (loser66):

bingo,

OpenStudy (loser66):

now, look back your problem, f(x) = ax +b and they ask you find f'(x) =?

OpenStudy (anonymous):

so in my question. f'(x)=a and g'(x)=c

OpenStudy (loser66):

WOAHH , you fired me??? hehehehe...

OpenStudy (loser66):

Congratulation, you got the correct answers

OpenStudy (anonymous):

so if I add the functions together then find the derivatives what happens

OpenStudy (loser66):

particular one, please. You take the few baby steps, so that I dare not to say something without seeing the problem.

OpenStudy (anonymous):

f(x)+g(x)=(ax+b)+(cx+d) find its derivative

OpenStudy (loser66):

Yes, exactly what we do, just add then take derivative. (the product and the quotient are not the same as sum and difference)

OpenStudy (anonymous):

(a+c)

OpenStudy (loser66):

yup

OpenStudy (anonymous):

if f(x) and g(x) are arbitrary functions what's the derivative of f(x)+g(x)?

OpenStudy (loser66):

if the degree of f and g is 1 as we did, then, do as we did if the degree of f and g is not 1, then, you must study another thing.

OpenStudy (anonymous):

Thank you so much. I will read on. I appreciate your help. Thank you!!!

OpenStudy (loser66):

for example, f(x) = 2x^3 -8x^2 +6 , you have to know how to take derivative of this function first. The degree of x is not 1

OpenStudy (loser66):

glad to help. good luck. Math is fun and beautiful. Win her!!

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