Question

Differentiate

Answer 1

\[f(x)= xe^xcscx\]

Answer 2

How would I break this up in order to use the rules?

Answer 3

Ooh that's an ugly one, product rule a couple of times

Answer 4

Using \[f'g+g'f\] first let f = x and g = e^xcscx and go from there

Answer 5

\[y ' = (1) \times (e^xcscx) + (x)(e^x \times (-cscxcotx))\]

Answer 6

\[-xe^xcscxcotx\]

Answer 7

\[-xe^x \times cscxcotx\]

Answer 8

Then use product rule for both sides?

Answer 9

I think we need to be a bit more careful here

Answer 10

\[\frac{ d }{ dx } (x)e^xcscx+\frac{ d }{ dx }(e^xcscx)x\] careful on what you're taking the derivative of

Answer 11

\[\frac{ d }{ dx }(x) = 1\] we get \[e^xcscx + \frac{ d }{ dx }(e^xcscx)x\] now do \[\frac{ d }{ dx }(e^xcscx)\]

Answer 12

\[\frac{ d }{ dx }(e^xcscx) = (e^x \times -cscxcotx) \times ( e^xcscx)\]
Using product rule.
= ?

Answer 13

This requires product rule

Answer 14

\[(fg)' = f'g+g'f\]

Answer 15

Mhm not quite

Answer 16

I think you meant + instead of x right :)

Answer 17

ahh, yes

Answer 18

\[\frac{ d }{ dx }(e^xcscx)\]
I did that wrong?

Answer 19

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

Answer 20

\[\frac{ d }{ dx }(e^x cscx) = e^xcscx -cotxcscxe^x\] I feel there is an easier way to approach this problem but lets keep going haha since I think you understand this.

Answer 21

okay

Answer 22

So we have \[e^xcscx+e^xcscx-cotxcscxe^x\] that's gross but you can simplify it xD

Answer 23

Oh I'm missing an x somewhere

Answer 24

\[e^xcscx+x(e^x)(cscx)-(x)(cotx)(cscx)e^x\]

Answer 25

I think that should suffice for your final answer

Answer 26

\[e^xcscx( 1 + x - xcotx)\] would be the simplified answer?

Answer 27

That should work

Answer 28

okay, thanks :)

Answer 29

Np