Calculate the gradient of the curve at the point where x = 1 ? y = x^3 +7

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Answer 1 · 2018-06-17T04:14:17

Gradient is #3#.

Gradient of a curve #y=x^3+7# is given by its derivative #(dy)/(dx)#.

As #y=x^3+7#, #(dy)/(dx)=3x^2#

And therefore gradient at #x=1# is #(dy)/(dx)=3*1^2=3*1=3#

Answer 2 · 2018-06-17T04:18:51

Slope of the curve at #(x=1)=3#

Given -

#y=x^3+7#

Slope of the curve is given by the first derivative.

#dy/dx=3x^2#

At #x=1# The slope of the curve -

At #x=1; dy/dx=3(1^2)= 3#

Slope of the curve at #(x=1)=3#