Question

Question #9c03c

Answer 1

`sec^2(2x+3) * 2`

Explanation:

`f(x) = tan(2x+3)`

The chain rule states that:

`(df(u))/dx = (df)/(du) * (du)/(dx)`

Setting `u = 2x+3`, we get the equation:

`(df(x))/dx = d/(du) tan(u) * d/(dx) 2x+3`

And since `d/(du) tan(u)` = `sec^2(u)` and `d/(dx) 2x+3` = 2:

`(df(x))/dx = sec^2(u) * 2`, which is:

`= sec^2(2x+3) * 2`

Answer 2

`dy/dx=2sec^2(2x+3)`

Explanation:

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

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

`dy/dx=f'(g(x))xxg'(x)`

`y=tan(2x+3)`

`rArrdy/dx=sec^2(2x+3)xxd/dx(2x+3)`

`color(white)(rArrdy/dx)=2sec^2(2x+3)`