Question

Find the derivative of `1+cosxsinx+3cos4x+7tan8x`?

Answer 1

`d/(dx)[1+cosxsinx+3cos4x+7tan8x]=cos2x-12sin4x+56sec^2 8x`

Explanation:

As `d/(dx)[a(x)+b(x)+c(x)+d(x)]=(da)/(dx)+(db)/(dx)+(dc)/(dx)+(dd)/(dx)`

`d/(dx)[1+cosxsinx+3cos4x+7tan8x]=0+d/(dx)sinxcosx+d/(dx)3cos4x+d/(dx)7tan8x` (as derivative of constant term is `0`)

Now we can use chain rule to find derivative of `cos4x` and

`d/(dx)3cos4x=3xxd/(d(4x))cos4x xxd/(dx)4x`

= `3xx(-sin4x)xx4=-12sin4x`

Similarly `d/(dx)7tan8x=7xxsec^2 8x xx8=56sec^2 8x`

To find derivative of `sinxcosx` we write it as `1/2sin2x` and

`d/(dx)sinxcossx=d/(dx)1/2sin2x=1/2xxcos2x xx2`

= `cos2x`

Hence `d/(dx)[1+cosxsinx+3cos4x+7tan8x]=cos2x-12sin4x+56sec^2 8x`

Answer 2

`y'=cos^2x-sin^2x-4(3sin(4x)-14sec^2(8x))`

`color(white)(X)=cos(2x)-4(3sin(4x)-14sec^2(8x))`

Explanation:

We can write `y` in terms of functions of `x`.

`y=1+f(x)g(x)+3h(x)+7j(x)`

`y'=f(x)g'(x)+f'(x)g(x)+3h'(x)+7j'(x)`

`f(x)=cosx`
`f'(x)=-sinx`

`g(x)=sinx`
`g'(x)=cosx`

`h(x)=cos(4x)`
`h'(x)=-4sin(4x)`

`j(x)=tan(8x)`
`j'(x)=8sec^2(8x)`

`y'=cosxcosx-sinxsinx+3(-4sin(4x))+7(8sec^2(8x))`

`=cos^2x-sin^2x-12sin(4x)+56sec^2(8x)`

`=cos^2x-sin^2x-4(3sin(4x)-14sec^2(8x))`

Using the double angle identity: `cos(2x)=cos^2x-sin^2x`, we get:

`=cos(2x)-4(3sin(4x)-14sec^2(8x))`