miss!!!! wat this question really tryn to say. Are u asking me to differentiate those functions. il differentiate it anyway. i tink u all shud learn this, since its where differentiation of exponentials, trig and ln functions all begin. if y = ln x dy/dx = 1/x
NOTE!!! To dirrerentiate an exponential function ie e^x..its just like normal differentiation. Consider this example. y = 4e^2x find dy/dx in this case we differentiate the power and multiply the differential by the coefficient and put back the remaining part of the function. so dy/dy = 8e^2x
cokebaby, : 1. y = 2e^2x. so dy/dx = (2)2e^2x = 4e^2x 2.y = 2sin 3x - 4cos 3x^2 for 2sin 3x: in the form y= a sinx u, so u =3x and du/dx = 3 y = sin x, so dy/dx = cos x i believe the constant remains the same.
WHY THE MASTER MUST ANSWER. I GETTING FED UP WITH THIS BLOG. but who cares coming to Moose questions, i will be happy to answer the "harder ones". question 1:
y = sin(e^3x)
let:
u = e^3x du/dx = 3e^3x
therefore; y = sin u dy/du = cos u
this implies:
dy/dx = dy/du * du/dx dy/dx = cos u * 3e^3x
dy/dx = (3e^3x)cos(e^3x) and here ends question 1.
question 2:
y = e^(sin x)
let: u = sin x du/dx = cos x
therefore:
dy/dx = (cos x)(e^(sin x))
i will love for anyone to ask me any related questions PLEASEEEE. i kind a bored nah so anytime u people want.
well, marz.: y = sin(x^2 + 3) take u = (x^2 + 3) du/dx = 2x dy/du = cos u dy/dx = dy/du x du/dx dy/dx = cos u x 2x dy/dx = 2xcos(x^2 + 3) i think this right. is this right .......anyone????...
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17 comments:
miss!!!! wat this question really tryn to say. Are u asking me to differentiate those functions.
il differentiate it anyway.
i tink u all shud learn this, since its where differentiation of exponentials, trig and ln functions all begin.
if y = ln x
dy/dx = 1/x
if y = e^x
dy/dx = e^x
if y = sin x
dy/dx = cos x
if y = cos x
dy/dx = - sin x
U NEED TO KNOW THOSE!!!
ORRRR!!! I GET IT.. I WAS SUPPOSE TO ASK QUESTIONS!!! OK!!!!
I start very simple.
To anyone!!!!! differentiate
a) y = 2e^-2x
b) y = 2sin3x - 4cos3x^2
NOTE!!! To dirrerentiate an exponential function ie e^x..its just like normal differentiation.
Consider this example.
y = 4e^2x
find dy/dx
in this case we differentiate the power and multiply the differential by the coefficient and put back the remaining part of the function.
so dy/dy = 8e^2x
cokebaby;
DIFFERENTIATE
question
y = sin(x^2 + 3).
dy/dx = ??????
question 2
y = cos 3x^4.
dy/dx = ?????
it starting easy bloggers don't worry harder questions are coming
y = sin(e^3x)
dy/dx = ?????????
BLOGBAKES
y = e^(sin x).
dy/dx = ??????????
can anyone answer the following questions?
1) y=cos2x + x^2
find dy/dx
2) y=sin3x^4 . 3x
find dy/dx
cokebaby, :
1. y = 2e^2x.
so dy/dx = (2)2e^2x
= 4e^2x
2.y = 2sin 3x - 4cos 3x^2
for 2sin 3x:
in the form y= a sinx u, so u =3x and du/dx = 3
y = sin x, so dy/dx = cos x
i believe the constant remains the same.
taz
y = cos2x + x^2
let u = 2x
dy/du = 2
y = cosu + x^2
dy/du= -sin u
v = x^2
dv/dx = 2x
dy/dx= -sin 2x + 2x
WHY THE MASTER MUST ANSWER. I GETTING FED UP WITH THIS BLOG. but who cares
coming to Moose questions, i will be happy to answer the "harder ones".
question 1:
y = sin(e^3x)
let:
u = e^3x
du/dx = 3e^3x
therefore;
y = sin u
dy/du = cos u
this implies:
dy/dx = dy/du * du/dx
dy/dx = cos u * 3e^3x
dy/dx = (3e^3x)cos(e^3x)
and here ends question 1.
question 2:
y = e^(sin x)
let:
u = sin x
du/dx = cos x
therefore:
dy/dx = (cos x)(e^(sin x))
i will love for anyone to ask me any related questions PLEASEEEE.
i kind a bored nah so anytime u people want.
AHHHHHHH SOMEONE WHO IS WILLING TO ANSWER QUESTIONS!!!!!!!!!!!!
THAT IS WHAT I WANT TO HERE. A CHALLENGE
y = sin(x^2 + 3)
dy/dx= cos x(2)2
dy/dx= 2 cosx(2)
hey can someone help me and correct me if im wrong cause i think i doing something wrong
But plz explain while you all show calculations it helps me much better
well, marz.:
y = sin(x^2 + 3)
take u = (x^2 + 3)
du/dx = 2x
dy/du = cos u
dy/dx = dy/du x du/dx
dy/dx = cos u x 2x
dy/dx = 2xcos(x^2 + 3)
i think this right. is this right .......anyone????...
ok cookie we all know you like to riight books
how do you know when to differentiate in terms of dy/dx??
it all depends on the unknown variables in the equation, like in the equation of y that contains x variables: use dy/dx.
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