Question

Derivative of sin(5x)*cos(x)^-1?

Answer 1

You can either rewrite it as:\[\frac{ 1 }{ \sin(5x)\cos(x) }\] an use the quotient then product rule. Or you can use the product and chain rule.

Answer 2

It wouldn't be written as \[\frac{ \sin(5x) }{ cosx }\]

Answer 3

oh is the -1 just attached the cosx? (it's hard to tell on here!) lol
in which case, yes you are correct, and if you just apply the quotient rule, are you familiar with the formula?

Answer 4

\[\frac{ \sin(5x) }{ \cos(x) }=\frac{ u }{ v } \quad \text{then} \quad \frac{ d }{ dx }=\frac{ v'u - uv' }{ v^2 }\]

Answer 5

\[\frac{ \cos(x)*5\cos(5x)-\sin(5x)(-\sin(x)) }{ \cos^2(x) }\]is what I got

Answer 6

Sorry I just realised I wrote the formula wrong (apologies it's early in the morning for me!) it should be: \[\frac{ vu'-uv' }{ v^2 }\] but you did it correctly anyway lol
You can further simplify this by splitting the equation into two parts:\[\frac{ \cos(x)5\cos(5x) }{ \cos^2(x) }+\frac{ \sin(5x)\sin(x) }{ \cos^2(x) }\] note it becomes positive due to the two minus signs. Can you simplify this further using sec/cosec?

Answer 7

I know the first part can be simplified down to\[\frac{ 5\cos(5x) }{ cosx }\]Not sure about the second part

Answer 8

okay, do you know that: \[\frac{ 1 }{ \cos(x) }=\sec(x)?\]

Answer 9

I know that

Answer 10

Okay, so you're okay with the fact that you can rewrite that first bit as:\[\frac{ 5\cos(5x) }{ \cos(x) }=5\cos(5x)*\frac{ 1 }{ \cos(x) }=5\cos(5x)\sec(x)\]

Answer 11

Okay, I got that. Can the second part be simplified?

Answer 12

First part sorted then!
Second part:\[\frac{ \sin(5x)\sin(x) }{ \cos^2(x) }=\sin(5x)*\frac{ \sin(x) }{ \cos(x)\cos(x) }=\sin(5x)\frac{ \sin(x) }{ \cos(x) }\frac{ 1 }{ \cos(x) }\] Can you simplify this for me?

Answer 13

\[\sin(5x)*\tan(x)*\sec(x)\]

Answer 14

that's it! all done!

Answer 15

so your final solution is: 5cos(5x)sec(x)+sin(5x)tan(x)sec(x)