Question

Can someone help me? the squareroot is throwing me off
Find the derivative of the function.
f(t) = (1 +sqrt t)(8t^2 − 3)

Answer 1

\[(1+\sqrt{t})(8t ^{2}-3)\]

Answer 2

I would write the sqr as t^(½) and use the power rule

Answer 3

I tried and still no good =/ could you show me how you'd solve this one as an example? I have the answer to it:

Problem: f(t) = (1 + t)(6t^2 − 3)

Answer: f '(t) = \[15t ^{3/2}+12t - 3/(2\sqrt(t)) \]

Answer 4

do you know the product rule?

Answer 5

that I have to multiply the whole number by the exponent and then subtract 1 from the exponent?

Answer 6

that is the "power rule"

the product rule is
d ( f * g) = f * g' + g * f'

if you don't know it we can do the problem another way.

Answer 7

hmm let me go over my instructional video again (online class) I saw that rule in it

Answer 8

you can review it later. Let's just use it
\[ \frac{d}{dt} \left((1+\sqrt{t})(8t ^{2}-3) \right) = \\
(1+\sqrt{t}) \frac{d}{dt}(8t ^{2}-3) + (8t ^{2}-3) \frac{d}{dt}(1+t^{\frac{1}{2}})
\]

Answer 9

next step, take the derivative of
\[ \frac{d}{dt}(8t ^{2}-3) \]
can you do that ?

Answer 10

16t?

Answer 11

yes. so far we have
\[(1+t^{\frac{1}{2}}) 16t + (8t ^{2}-3) \frac{d}{dt}(1+t^{\frac{1}{2}}) \]

if we distribute the 16t in the first term, do you know what we get ?

Answer 12

btw, I notice we have a discrepancy in the problem. We probably want to be doing
Problem: f(t) = (1 + t)(6t^2 − 3)

NOT Problem: f(t) = (1 + t)(8t^2 − 3)

that changes your derivative to 12 t (not 16 t)

Answer 13

Ok then it'd be 12t + 16t^3/2

Answer 14

12t^3/2

Answer 15

ok, so far so good. now the other derivative:
\[ \frac{d}{dt}(1+t^{\frac{1}{2}}) \]

Answer 16

this website really needs to fix these problems, every time I reply I have to sign back in and find the question sorry I take so long! and ok would that be 1/2t^-1/2 ?

Answer 17

you should not have to sign back in. but meanwhile we have
\[ 12t + 12 t^{\frac{3}{2}} + (6t^2 − 3)\frac{1}{2} t^{-\frac{1}{2}} \]

Answer 18

12t + 12t^3/2 +3^-1 - 1.5t^-1/2
\[12t+12t ^{3/2}+3^{-1}-1.5^{-1/2}\]

Answer 19

distribute, collect common terms, and re-write t^(-½) as 1/sqrt(t)
and you get your answer.

Answer 20

?
\[ 6 t^2 \cdot \frac{1}{2} \cdot t^{-\frac{1}{2}} = \frac{6}{2} \cdot t^2 \cdot t^{-\frac{1}{2}} \\
= 3 t^{2-\frac{1}{2}} = 3 t^{\frac{3}{2} }
\]

Answer 21

and because the answer does not use decimals, I would leave it 3/2

Answer 22

Sorry bout that I read 6t for some reason

Answer 23

can you finish ?

Answer 24

Yeah thanks so much, it's 15t^3/2+12t-3/2sqrt(t), the answer :)