Question

Question #648c3

Answer 1

`f'(x)=6tan3xsec^2 3x`

Explanation:

We have
`f (x)=1/cos^2(3x)`
Differentiating both sides with respect to `x` using chain rule we get
`f'(x)= -2/cos^3 (3x)×(-sin3x)×3`
`=6tan3xsec^2 3x`

Answer 2

`6tan3xsec^2 3x`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"given "y=(f(g(x))" then"`

`dy/dx=f'(g(x)xxg'(x)larr" chain rule"`

`y=1/(cos^2 3x)=(cos3x)^(-2)`

`rArrdy/dx-2(cos3x)^(-3)xxd/dx(cos3x)`

`color(white)(y)=-2(cos3x)^(-3)xx-sin3x xxd/dx(3x)`

`color(white)(y)=(6sin3x)/(cos3x)^3`

`color(white)(y)=6tan3x xx1/(cos 3x)^2`

`color(white)(y)=6tan3xsec^2 3x`