At $P(x_1,y_1)$, differentiate before substituting $x_1$. \(m=\mathrm{f}^{\prime}(x_1)\qquad\tex

Question

At $P(x_1,y_1)$, differentiate before substituting $x_1$. $m=\mathrm{f}^{\prime}(x_1)\qquad\text{Tangent:}\hspace{0.5em}y-y_1=m(x-x_1)$ $\text{Normal:}\hspace{0.5em}y-y_1=-\dfrac1m(x-x_1)\quad(m\ne0)$ 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. Example Find the gradient of the curve $y=(x-1)(2x+3)$ at $x=2$. $y$${}=2x^2+3x-2x-3$${}=2x^2+x-3$ $\dfrac{\mathrm{d}y}{\mathrm{d}x}=4x+1$ At $x=2$, $\text{gradient}$${}=4(2)+1$${}=9$

Accepted Answer

$9$

Gradient of Curve

Gradient of Curve

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