What is the equation of the line tangent to # f(x)=e^(x^2+x) # at # x=0 #?

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Answer 1 · 2018-04-27T09:49:25

#y-x-1=0#

To find the point (0,y) which lies on the tangent We substitute in the function with the value #x=0#

#f(0)=e^(0+0)=1#

So the point is # (0,1)#

to find the slope#(m)# of the tangent we find the first derivative #((dy)/dx)# value at the point #(0,1)#

#(dy)/dx=e^(x^2+x)(2x+1)## "rarr#Chain Rule

Substitute with the point #(0,1)#

#dy/dx=e^(0+0)*(2xx0+1)=1#

The Equation of a Straight Line

#color(green)((y-y_1)=m(x-x_1)#

Substitute

#y-1=1(x-0)#

#y-x-1=0#