Question

Find the point on which curve y=x^3,where the tangent is inclined at an angle of 60degree with x-axis​

Answer 1

Step-by-step explanation:

remember that slope is the tan of angle of inclination which is dy/dx at that point

Answer 2

Differentiating,

dy/dx=3x^2=tan 60°= $\sqrt{3}$

x^2=1/$\sqrt{3}$

x=1/$\sqrt{\sqrt{3} }$

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