Is this right? (1 Viewer)

[abdog]

Differentiate sinx(3x+1).

u=sin x
du=cos x
v=3x+1
dv=3

Therefore, dy=3sinx + 3xcosx + cosx

 

[yasminee96]

Yep
alternative (not much difference):

expand it
so
sinx(3x+1)
= 3xsinx + sinx
then differentiate.

just makes it less confusing

 

[braintic]

The question is ambiguous. Its not clear whether it means (3x+1)sinx or sin[x(3x+1)]

 

[Nws m8]

The question is ambiguous. Its not clear whether it means (3x+1)sinx or sin[x(3x+1)]

+1

 

[abdog]

It's the former

 

[abdog]

But for some reason, the answer is:

3cos(3x+1)

 

[Shadowless]

Then shouldn't it be:

f(x) = sin(3x+1)

?

 

[abdog]

Nope, the question is sinx(3x+1). Also, how do you simplify sinx=cosx? Does it become sinx/cosx = 1?

 

[Shadowless]

hmm... well if...

f(x) = (sin[x]).(3x+1)

then...

f'(x) = (sin[x])(3) + (3x+1)(cos[x])
= 3sin[x] + 3xcos[x] + cos[x]

?

In relation to your statement 'does it become (sin[x])/(cos[x]) = 1', the answer is no.

It's... 'sin[x]/cos[x] = tan[x]'.

 

[abdog]

So I got a question that says find the point of intersection between y=cosx and y=sinx. So sinx=cosx. How do I go from here?

 

[deswa1]

Divide both sides by cosx

 

[abdog]

Wow, never thought of that before, thanks!