Question

Question #ad2ca

Answer 1

`f'(x)=(sinxcosx-x(cos^2x-sin^2x))/(sinxcosx)^2`

`=(2(1-2xcot(2x)))/sin(2x)`

Explanation:

`f(x)=x/(sinxcosx)=g(x)/(h(x))`

`g(x)=x`
`g'(x)=1`

`h(x)=sinxcosx`
`h'(x)=cosxcosx-sinxsinx=cos^2x-sin^2x`

Using the double angle formula, `h'(x)` can be simplified down to `cos(2x)`

`f'(x)=(h(x)g'(x)-h'(x)g(x))/(h(x))^2`

`=(sinxcosx-x(cos^2x-sin^2x))/(sinxcosx)^2`

Continuing with the double angle rules:

`=(sinxcosx-xcos(2x))/(sinxcosx)^2`

`=1/(sinxcosx)-(xcos(2x))/(sinxcosx)^2`

`sinxcosx=1/2(2sinxcosx)=1/2sin(2x)`

`f'(x)=1/(1/2sin(2x))-(xcos(2x))/(1/2sin(2x))^2`

`=2/sin(2x)-(xcos(2x))/(1/4sin^2(2x))`

`=2/sin(2x)-(4xcos(2x))/(sin^2(2x))`

`=2/sin(2x)-(4xcot(2x))/(sin(2x))`

`=(2-4xcot(2x))/sin(2x)`

`=(2(1-2xcot(2x)))/sin(2x)`

Answer 2

`color(blue)(1/(Cos x Sin x)-x/(Sin^2 x) + x/Cos^2 x)`

Explanation:

We are given a function of `x` that we must differentiate:

`f(x) = [x/(Sin x Cos x) ]` `color(red)(Function.1)`

`rArr f(x) = y = [x/(Sin x Cos x) ]`

We need to find the First Derivative of `f(x)`

By observing `f(x)` we know that we must use the Quotient Rule to differentiate .

Quotient Rule for finding the derivatives states that

`color(blue)((dy)/(dx)[f(x)/g(x)] = [(g(x)*f'(x) - f(x)*g'(x))/[g(x)]^2]`

Using the Quotient Rule, we can write our `color(red)(Function.1)` as

`=[d/(dx)[x]*[Cos x Sin x] -x*d/(dx) [ Cos x Sin x]]/[Cos x Sin x]^2`

Product Rule for finding the derivatives states that

`color(blue)((dy)/(dx)[f(x)*g(x)] = [(f'(x)*g(x) + f(x)*g'(x)]`

`=[1*[Cos x Sin x] -x*[d/(dx) [ Cos x]* Sin x+Cos x*d/(dx)[Sin x]]]/[Cos^2 x Sin^2 x]`

On simplification we get,

`=[[Cos x Sin x] -x[- Sin x* Sin x+Cos x Cos x]]/[Cos^2 x Sin^2 x]`

We can simplify further to get

`=[[Cos x Sin x] -x[- Sin^2 x+Cos^2 x]]/[Cos^2 x Sin^2 x]`

We can rearrange terms to get

`=[[Cos x Sin x] -x[Cos^2 x - Sin^2 x]]/[Cos^2 x Sin^2 x]`

We can simplify further to obtain

`=[Cos x Sin x]/[Cos^2 x Sin^2 x] -[x[Cos^2 x - Sin^2 x]]/[Cos^2 x Sin^2 x]`

We can rewrite the above expression as

`=[Cos x Sin x]/[[Cos x Sin x]*[Cos x Sin x]] -[x*Cos^2 x]/[Cos^2 x Sin^2 x] + [x*Sin^2 x]/[Cos^2 x Sin^2 x]`

We can now cancel terms as

`=[cancel(Cos x Sin x)]/[[cancel(Cos x Sin x)]*[Cos x Sin x]] -[x*cancel(Cos^2 x)]/[cancel(Cos^2 x) Sin^2 x] + [x*cancel(Sin^2 x)]/[Cos^2 x cancel(Sin^2 x)]`

In this step, we get

`1/(Cos x Sin x)-x/(Sin^2 x) + x/Cos^2 x` ....... Final Result

I hope you find this solution helpful.