Find the gradient of y=6x^3+2x^2 at (1,1)
In order to find the gradient of the curve at (1,1), we must first differentiate the equation of the curve. To do this, multiply the coefficient of x by the power of that same x. Then subtract one from the power. (d/dx)(6x^3)=(36)x^(3-1)=18x^2. While (d/dx)(2x^2)=(22)x^(2-1)=4x. Therefore, the derivative of the equation is (dy/dx)=18x^2+4x.
To find the gradient of the equation at (1,1), substitute x=1 into the derivative. 18(1)^2+4(1)=22. So the gradient of y=6x^3+2x^2 at (1,1) is 22.
N.B. In tutorials I will use a whiteboard for my workings.
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