find the gradient of the tangent to the curve y=x^2 at the point (4,16)
First of all you must differentiate this function, as this will give you the gradient function necessary to find the gradient at the point (4,16). To differentiate it you use the formula dy/dx= anx(n-1), where a is the coefficient of the x term, and n is the power it has been raised to. So in the case y=x2 n=1 (as there are no coefficients of x2) and n=2 as x has been raised to the power of 2. Therefore dy/dx= 12x(2-1) = 2x.Now we have our gradient function, which is dy/dx=2x, we can simply plug in the x value of the point on the curve we want to find the gradient of, in this case x=4. Therefore at the point (4,16), the gradient of the curve is 2*(4) = 8.If you imagine drawing a straight line that is exactly in line with the trajectory of the curve at the point (4,16), this is known as a tangent to the curve at that point. As the tangent is exactly in line with the curve at that point, its gradient will be the same as the gradient of the curve. Therefore the gradient of the tangent to the curve at the point (4,16) is 8.
