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:

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