Question

Can you apply product rule to :
\[\Large f(x)=e^{x}.sin(x)+e^{x}.cos(x)\]

Answer 1

Yes..

Answer 2

You should first take e^x common:

\[f(x) = e^x(\sin(x) + \cos(x))\]

\[f'(x) = e^x(\cos(x)-\sin(x)) + (\sin(x) + \cos(x)).e^x\]

Answer 3

Is your answer is like that ? how we applied it...
\[f^{(2)}(x)=e^{x}.sin(x)+e^{x}.cos(x)+e^{x}.cos(x)-e^{x}.sin(x)\]

Answer 4

Yes this is absolutely correct..

Answer 5

but i dont get how we used this rule here \[f^{'}(x)=u^{'}v+uv^{'}\]

Answer 6

See, you can write the given function as:

\[f(x) = e^x(sinx + cosx)\]

Till here you got??

Answer 7

yes

Answer 8

Here, \(u = e^x\) and \(v = (sinx + cosx)\)

Answer 9

ok..

Answer 10

So,

\[f'(x) = e^x \frac{d}{dx}(sinx + cosx) + (sinx + cosx) \frac{d}{dx}(e^x)\]

Right till here??

Answer 11

hmm i get confused..

Answer 12

the function of \[\frac{d}{dx}\] is not so clear to me

Answer 13

This is according to your definition..

See,

\[f'(x) = u.v' + v.u'\]

\(\frac{d}{dx}\) It is called derivative..

Answer 14

ok..

Answer 15

\[\frac{d}{dx}(sinx + cosx) = \frac{d}{dx}sinx + \frac{d}{dx}cosx = cosx - sinx\]

Answer 16

and:

\[\frac{d}{dx}(e^x) = e^x\]

It remains same..

Answer 17

ok i got here..

Answer 18

and one step above..

Answer 19

\[\frac{d}{dx}sinx = cosx\]

\[\frac{d}{dx}cosx = -sinx\]

Answer 20

thats ok..

Answer 21

maybe i get it now...

Answer 22

So, substitute these values in that function:

\[f'(x) = e^x(cosx - sinx) + (sinx + cosx).e^x\]

Answer 23

so just distribute e^x to the brackets you will get your answer..

Answer 24

ok thank you very much i think i am so near to undrstand,

Answer 25

Concentrate you will surely get this..