Question

Differentiate the function.
g(t) = 4t^−1/8

Answer 1

Power rule states
\[ \frac{d}{dx} (kx^n) = knx^{n-1}\]
Where k and n are Real constants. So what is k and n in your case?

Answer 2

uh... I'm going to say -1/8?

Answer 3

and 4?

Answer 4

see what i think you would do to find the answer is multiply 4 times (1/8) and then subtract 1 from the exponent

Answer 5

That is correct. Put all those things together and you have your answer

Answer 6

but when i put them together the answer doesn't make sense

Answer 7

i get -.6250538953 which can't be right

Answer 8

One moment.

Answer 9

lol

Answer 10

\[g(t) = 4t^{−1/8}\]\[g'(x)=-\frac{1}{8}\times 4x^{-\frac{1}{8}-1}\] is a start

Answer 11

then
\[g'(x)=-\frac{1}{2}x^{-\frac{9}{8}}\]

Answer 12

power rule right?

Answer 13

when i plug that into my calc is says domain error

Answer 14

yeah

Answer 15

wait how did you get 9/8?

Answer 16

the fractions are confusing me

Answer 17

like i said in the last post
the hardest part of calc one is adding and subtracting fractions

Answer 18

when i plug the answer you gave me on my online homework it says its incorrect

Answer 19

btw i got \(-\frac{9}{8}\)
\[-\frac{1}{8}-1=-\frac{1}{8}-\frac{8}{8}=-\frac{9}{8}\]

Answer 20

did you plug in 9/8 or -9/8 for the exponent ?

Answer 21

-9/8

Answer 22

and this is the function right? \[g(t) = 4t^{−1/8}\]

Answer 23

\[4t ^{-1/8}\]

Answer 24

is the equation

Answer 25

then i will stick with
\[g'(t)=-\frac{1}{2}t^{-\frac{9}{8}}\]

Answer 26

I'm got the same exact answer i don't understand why its not working. this is why i hate online math!

Answer 27

Can you take a screenshot of how you entered it?

Answer 28

sometimes i am wrong, but this one i would bet \(\$1,000\) on
make sure to use a \(t\) as the variable, maybe that was the mistake

Answer 29

how do you take a screenshot on a macbook pro?

Answer 30

there are many reasons to hate on line math that is for sure
webassign? mymathlabs?

Answer 31

printscreen button i think says maybe prtscr

Answer 32

webassign!

Answer 33

I believe its (⌘)-Shift-3. They should be saved on yoru desktop

Answer 34

Press Command (⌘)-Shift-3.

Answer 35

Thats the mac button, the clover + shift-3

Answer 36

use a \(t\) not an \(x\)

Answer 37

omg wow! thank you.

Answer 38

im so sorry!

Answer 39

no it was my fault

Answer 40

i inadvertently switches to x's cause that is what i am used to

Answer 41

calc one on line?

Answer 42

no i would of done the same i tried it before like that and it didn't work so don't worry about it and yeah the homework is.

Answer 43

mind helping with one more? I'm sure you do the same with the exponent and everything \[f(x)=\frac{ 2 }{ 5 }x ^{10}\]

Answer 44

same damned thing

Answer 45

\[f'(x)=4x^9\] now it is easier, not fractions

Answer 46

i'm thinking you multiply the fraction and subtract 1 from the exponent

Answer 47

yeah
\[10\times \frac{2}{5}=4\]

Answer 48

okay i just wanted to make sure i was right so it wouldn't lock me out!

Answer 49

lol good
more?

Answer 50

i have a LOT lol do you know anything about differentiate

Answer 51

Differentiate \[g(x)= 3e ^{x}\sqrt{x}\]

Answer 52

that is what you are doing differentiating

Answer 53

I know x square is equivalent to x1/2

Answer 54

product rule for this one
\[(fg)'-f'g+g'f\] with
\[f(x)=e^x,f'(x)=e^x,g(x)=\sqrt{x}, g'(x)=\frac{1}{2\sqrt{x}}\]

Answer 55

typo there are on the first line, i meant \[(fg)'=f'g+g'f\]

Answer 56

yeah you are right about x^2 but let me give you a small piece of advice

Answer 57

okay whats the advice lol, i need all i can get

Answer 58

the square root is a very very common function
math teachers love to put it on quizzes and exams
while your colleages are differentiating it by writing
\[\sqrt{x}=x^{\frac{1}{2}}\\
\frac{d}{dx}[x^{\frac{1}{2}}]=\frac{1}{2}x^{\frac{1}{2}-1}\] etc, just memorize it , it never changes

Answer 59

okay

Answer 60

\[\frac{d}{dx}[\sqrt{x}]=\frac{1}{2\sqrt{x}}\] and that is all
like knowing \(7\times 8=56\) just know it

Answer 61

okay so then what is our first step in solving this problem

Answer 62

knowing that it is a product and that you have to use the product rule

Answer 63

\[\large (fg)'=f'g+g'f\]

Answer 64

so we multiply it by the same thing

Answer 65

in this case \[f(x)=e^x,f'(x)=e^x,g(x)=\sqrt{x}, g'(x)=\frac{1}{2\sqrt{x}}\]

Answer 66

leave the 3 out front as a common factor, and them make a direct substitution into the formula i just wrote above

Answer 67

\[3\left(e^x\sqrt{x}+e^x\times \frac{1}{2\sqrt{x}}\right)\] is the first step

Answer 68

and then we multiply by 3 for each term?

Answer 69

then clean up with some algebra, probably write
\[3\left(\sqrt{x}e^x+\frac{e^x}{2\sqrt{x}}\right)\] you can distribute if you like,
i would leave it like that

Answer 70

you get more or less what i did? a direct substitution into the "product rule"

Answer 71

why did you switch around the e^x's

Answer 72

i get the multiplying but I'm still confused

Answer 73

no real reason, just generally you see it written that way
\[xe^x\] for example rather than \[e^xx\] multiplication is commutative so it makes no real difference,

Answer 74

you want a more detailed explanation? i can try if you like

Answer 75

well yes i understand the xe^x but why did you replace the 1 with e^x

Answer 76

oooh i see your question

Answer 77

at first i wrote the last term as
\[e^x\times \frac{1}{2\sqrt{x}}\] then i confused you by writing it as
\[\frac{e^x}{2\sqrt{x}}\]

Answer 78

oh no that makes sense i get it now

Answer 79

but it arithmetic with fractions is all
just like
\[5\times \frac{1}{3}=\frac{5}{3}\]

Answer 80

you get how the product rule works? we can do another if you like

Answer 81

you multiple f into g and g into f

Answer 82

I have this one that I need help solving Differentiate. (Assume c is a constant.)

Answer 83

f primes times g plus g prime times f

Answer 84

\[z=x ^{3/2}(x+ce ^{x})\]

Answer 85

would we do the exponent thing?

Answer 86

for this i would multiply out first

Answer 87

\[\large z=x^{\frac{5}{2}}+cxe^x\] would be my first step

Answer 88

how did you get the 5/2?

Answer 89

i added one to the exponent

Answer 90

that is not the derivative !

Answer 91

i was just multiplying out
\[\large x^{\frac{3}{2}}\times x=x^{\frac{5}{2}}\]

Answer 92

maybe it would be easier not to do that and go right to the product rule
lets try that way

Answer 93

\[z=x ^{3/2}(x+ce ^{x})\]
\[z'=\frac{d}{dx}x^{3/2}(x+ce^x)+x^{3/2}\frac{d}{dx}(x+ce^x)\]

Answer 94

from there we get
\[z'=\frac{3}{2}x^{1/2}(x+ce^x)+x^{3/2}(1+ce^x)\] and that is all, i wouldn't change it

Answer 95

thank you

Answer 96

yw
hope the steps were clear