How do you find the equation of tangent line for the equation: #y = x^2 + sin(x)# at #x = pi#?

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Answer 1 · 2015-04-30T14:47:00

Find the equation of tangent line for the equation: #y = x^2 + sin(x)# at #x = pi#

We need a point and a slope.

At #x=pi#, the #y# value on the curve is:
#y=(pi)^2+sin(pi) = pi^2 +0 = pi^2#

So we have the point #(pi, pi^2)#

To find the slope, we'll use the derivative:

#y' = 2x +cos(x)# so at the point where #x= pi#, the slope on the tangent line to the curve is:

#m = 2(pi)+cos(pi) = 2pi -1#

Now use your favorite method for finding the equation of the line through #(pi, pi^2)# with slope #m = 2pi -1#.

Here's my solution using point-slope form: #y-y_1=m(x-x_1)#

#y-pi^2 = (2pi -1)(x-pi)#

Ok that is an equation for the line, but many prefer slope intercept form, so we'll solve for #y#.

#y-pi^2 = (2pi -1)x-(2pi-1)pi #

#y-pi^2 = (2pi -1)x - 2pi^2 +pi #. So

#y = (2pi -1)x - pi^2 +pi #.

Some might prefer adding parentheses around the #y#-intercept:

#y = (2pi -1)x +(- pi^2 +pi) #.