Question

I can't seem to get the correct answer for solving this derivative using the product rule...

Answer 1

\[h (t) = \sqrt{t} (1 - t ^{2})\]

Answer 2

\[h \prime (t) = (\sqrt{t}) (-2t) + ( 1 - t ^{2}) (1/2 t ^{-1/2})\]

This is what I got after plugging in the product rule

Answer 3

looks good

Answer 4

Arey ....

\[h(t) = \sqrt{t} - t^{5/2} \]\[\frac{d h(t)}{dt} = \frac{1}{2} \times t^{1/2 -1 } + \frac{5}{2} \times t ^{\frac{5}{2} - 1} \]\[h'(t) = \frac{1}{2 \times \sqrt{t}} + \frac{5 t^{3/2}}{2}\]

Answer 5

Where did the \[-t ^{5/2}\] come from?

Answer 6

t^1/2 x t^ 2 = t^1/2+2 = t^5/2

Answer 7

Oh okay, I didn't know you could combine \[\sqrt{t} , and -t ^{2}\] like that. I'll have to keep that in mind. Thanks

Answer 8

remember a(b+c)=ab+ac

Answer 9

but what you had done above using product rule was fine too

Answer 10

myininaya do you know how can we drive derivative of inverse functions

Answer 11

should i post it as a question there you can tell me

Answer 12

yes i think i remember how ok ishaan give me a few i would rather write on paper and scan and post

Answer 13

thanks

Answer 14

wrong attachment

Answer 15

thanks ...I tried it with limits and all... got confused ...thank you very much : )

Answer 16

the trick is too draw a picture
well some peeps might not need a picture but i certainly do

Answer 17

i know you probably get it by now so i will stop lol
its kindof fun though

Answer 18

yeah i got it thanks again it's a good solution : )

Answer 19

i derived some of inverse trig things above

Answer 20

let me know if you have any questions,chaise

Answer 21

Thanks a bunch! I'm afraid I'm not aware of what csc and sec are. Should I know what these are?

I don't want to burden by asking, cause I understand how difficult it can be to explain, but I will try and derive as much knowledge from your examples and also ask my maths teacher to further explain.

Answer 22

the trick is to make a right traingle and say we are wanting to find the derivative of y=arcsin(x)

then taking sin of both sides we have
siny=x

so y is an angle and we have siny=x/1 so that means the opposite side of that angle has measure x and the hyp. has measure 1
so we can find the other side by using Pythagorean thm
sqrt{1^2-x^2}=sqrt{1-x^2}
so we have siny=x
we can use implicit differentiation to find y'
cosy*y'=1
y'=1/cosy
but what is this in terms of x
well we can referred to our triangle right?
cosy means adj/hyp
hyp=1
and we found the adjacent side earlier using the Pythagorean thm

Answer 23

sec means hyp/adj

csc means hyp/opp

Answer 24

cos means adj/hyp

so you can say sec means 1/cos

Answer 25

These new terms are so foreign to me. Why have I never be explained these before?

Is there one your missing? Like.. a tan function one?

Answer 26

yes you can also do the above process to find the derivative of tan inverse
and csc inverse

Answer 27

cool. thank you so much!

Answer 28

if we have y=arctan(x), then tany=x right?
tany means opposite/adjacent right?
we have tany=x/1
so what side is missing and what is its measure?

Answer 29

we are missing the opposite side and its measure is 1/tany?
Is this right?

Answer 30

no ok tany means opposite/adjacent right?

Answer 31

so we are missing the hypotenuse

Answer 32

tan(y) = x/1
tan y = opp/adj?

The opposite side is x and the adj is 1?

Answer 33

Oooh. we are missing hyptonusel but the x is unknown. I see now. sorry.

Answer 34

| \
| \ h
x| y\
---
1
right?
so we need to know what h is
we can find measure h by using pythagorean thm
h=sqrt{x^2+1}

Answer 35

yes. im following.

Answer 36

ok so anyways we will come back to the triangle
we have tany=x right?
so taking derivative of both sides we get
y' sec^2(y)=1

Answer 37

to find y' which is our objective
we divide both sides by sec^2(y)
y'=1/(sec^2y)

Answer 38

but we need the derivative in terms of x

Answer 39

using our traingle what is secy in terms of x

Answer 40

Aah. and then you have to change the derivative back to terms of x, and yes I see now. :)
THANK YOU SO MUCH.