Question

What is the slope of the line tangent to the graph of the function `f(x)=ln(sin^2(x+3))` at the point where `x=pi/3`?

Answer 1

See below.

Explanation:

If:

`y=lnx<=>e^y=x`

Using this definition with given function:

`e^y=(sin(x+3))^2`

Differentiating implicitly:

`e^ydy/dx=2(sin(x+3))*cos(x+3)`

Dividing by `e^y`

`dy/dx=(2(sin(x+3))*cos(x+3))/e^y`

`dy/dx=(2(sin(x+3))*cos(x+3))/(sin^2(x+3))`

Cancelling common factors:

`dy/dx=(2(cancel(sin(x+3)))*cos(x+3))/(sin^cancel(2)(x+3))`

`dy/dx=(2cos(x+3))/(sin(x+3))`

We now have the derivative and will therefore be able to calculate the gradient at `x=pi/3`

Plugging in this value:

`(2cos((pi/3)+3))/(sin((pi/3)+3))~~1.568914137`

This is the approximate equation of the line:

`y=15689/10000x-1061259119/500000000`

GRAPH: