find the slope of the tangent line to f(x)=x2 + 2x at x=3
Solution:
Given: f(x)=x2+2xf(x)=x^2+2xf(x)=x2+2x ...(i)
Slope of tangent is f′(x)f'(x)f′(x)
So, we find derivative of given f(x)f(x)f(x)
On differentiating (i) w.r.t xxx
f′(x)=2x+2f'(x)=2x+2f′(x)=2x+2
Putting x=3x=3x=3 in f′(x)f'(x)f′(x) , we get,
f′(3)=2(3)+2=6+2=8f'(3)=2(3)+2=6+2=8f′(3)=2(3)+2=6+2=8
Answer: 8
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