How do you find the tangent line of #f(x) = 3xsinx^2 # at x=-1?

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Answer 1 · 2015-12-23T17:59:39

#y+2.254=5.766(x+1)#

#f(x)=3xsin(x^2)#

Use the product rule to find #f'(x)#.

#f'(x)=sin(x^2)d/dx[3x]+3xd/dx[sin(x^2)]#

#=>3sin(x^2)+6x^2cos(x^2)#

#f(-1)=-3sin(1)~~-2.524#

Thus, the tangent line passes through the point #(-1,-2.524)#.

#f'(-1)=3sin(1)+6cos(1)~~5.766#

The slope of the tangent line is #5.766#.

Write this in point-slope form:

#y+2.254=5.766(x+1)#