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 $x=pi/3$ ?

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Answer 1 · 2018-03-14T15:03:57

See below.

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:

enter image source here

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