At P(x_1,y_1), differentiate before substituting x_1. m=f (x_1): y-y_1=m(x-x_1) Normal: y-y_1=- 1m(x-x_1) (m 0) If m=0, the normal is x=x_1. A tangent may cross the curve, as at the origin of y=x3. If the gradient is given, solve f (x)=m to locate the point. Example Find the gradient of the curve y=(x-1)(2x+3) at x=2. y =2x2+3x-2x-3 =2x2+x-3 dydx=4x+1 At x=2, gradient =4(2)+1 =9

At \(P(x_1,y_1)\), differentiate before substituting \(x_1\).

\(m=\mathrm{f}^{\prime}(x_1)\qquad\text{Tangent: }y-y_1=m(x-x_1)\)

\(\text{Normal: }y-y_1=-\dfrac1m(x-x_1)\quad(m\ne0)\)

If \(m=0\), the normal is \(x=x_1\). A tangent may cross the curve, as at the origin of \(y=x^3\).

If the gradient is given, solve \(\mathrm{f}^{\prime}(x)=m\) to locate the point.

Find the gradient of the curve \(y=(x-1)(2x+3)\) at \(x=2\).

\(y\)\({}=2x^2+3x-2x-3\)\({}=2x^2+x-3\)

\(\dfrac{\mathrm{d}y}{\mathrm{d}x}=4x+1\)

\(\text{gradient}\)\({}=4(2)+1\)\({}=9\)

Differentiation

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Gradient of Curve

Gradient of Curve

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