What is the equation of the line tangent to #f(x)=x^2 + sin^2x # at #x=pi#?

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Answer 1 · 2016-03-15T09:09:08

Find the derivative and use the definition of the slope.

The equation is:

#y=2πx-π^2#

#f(x)=x^2+sin^2x#

#f'(x)=2x+2sinx(sinx)'#

#f'(x)=2x+2sinxcosx#

The slope is equal to the derivative:

#f'(x_0)=(y-f(x_0))/(x-x_0)#

For #x_0=π#

#f'(π)=(y-f(π))/(x-π)#

To find these values:

#f(π)=π^2+sin^2π#

#f(π)=π^2+0^2#

#f(π)=π^2#

#f'(π)=2*π+2sinπcosπ#

#f'(π)=2*π+2*0*(-1)#

#f'(π)=2π#

Finally:

#f'(π)=(y-f(π))/(x-π)#

#2π=(y-π^2)/(x-π)#

#2π(x-π)=y-π^2#

#y=2πx-2π^2+π^2#

#y=2πx-π^2#