Question

Derivative of:
f(x) = (e^7x)(sin(8x))
g(x) = (x^6)(sin(9x))
h(t) = 8(tan(t^6))

Answer 1

product rule and chain rule will get you thru it

Answer 2

if there is a specific question you have about it, just ask, otherwise this just seems to come across as you wanting a free answering service

Answer 3

f(x) = (e^7x)(sin(8x))
f'(x) = (e^7x)(sin8x)' + (sin8x)(e^7x)'

for (d/dx)(sinnx)=dsinnx/dnx = ncosnx
for d/dx (e^nx) = de^nx/dnx = ne^nx

you can then find the ans

Question (2) is similar to quesiton (1)

h(t) = 8(tan(t^6))
dtanx/dx=sec^x
dt^ndt = nt^(n-1)
then, you can get the ans

Try it yourself

Answer 4

The answers I got were: F'(x) = 56(e^(7x))(cos(8x)); g(x) = 54(x^5)(cos(9x)); and h(t) = 6(t^5)(sec(t^6))^2... and apparently they're incorrect. I'm well aware amistre that I have to do product and chain rule and have attempted that... you don't have to be obnoxious to people looking for a little help.

Answer 5

amendment : dtanx/dx=sec^2 x

Answer 6

actually, i do in order to evaluate your responses and intention in order to police the site appropriately.

Answer 7

f'(x) = (e^7x)(sin8x)' + (sin8x)(e^7x)'
= 8e^7x cos 8x <--- dsinx=cosx (don't forget to d8x which is inside sin)
+ 7sin8x e^7x <--- de^7x =7e^7x (don't forget d7x=7)
=e^7x( 8cos8x+7sin8x) <--- take out the common factor

Answer 8

g(x) = (x^6)(sin(9x))
g'(x) = (x^6)'(sin(9x))+(x^6)(sin(9x))'
=6x^5 sin9x +9 x^6 cos9x
= 3x^5 ( 2sin9x+3xcos9x)
This is similar to f'(x)

Answer 9

oh okay.. tnx CoCoTsoi.. I see what I missed.. thanks again

Answer 10

I got it from here

Answer 11

h(t) = 8(tan(t^6))
h'(t) = 8 sec^2 (t^6) 6t^5
= 48t^5 sec^2 (t^6)

Answer 12

to mirror coco

[fg]' = f'g+fg'

f(x) = (e^7x)(sin(8x))

f'(x) = (e^7x)'(sin(8x)) + (e^7x)(sin(8x))'


f'(x) = 7(e^7x)(sin(8x)) + 8(e^7x)(cos(8x))

f'(x) = e^7x(7sin(8x) + 8cos(8x))

Answer 13

I am glad to help you :D