<p>For a curve with equation\(y=\mathrm{f}(x)\),<br>Gradient for tangent, \(m=\frac{\mathrm{d}y}{\mathrm{d}x}\) <br>Gradient for normal, \(-\frac{1}{m}=-\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}\) (since, \({{m}_{1}}\times {{m}_{2}}=-1\) )</p>

For a differentiable curve \(y=\mathrm{f}(x)\), the derivative gives the tangent gradient at the point under consideration.

The two non-zero finite gradients satisfy \(m_{\mathrm{t}}m_{\mathrm{n}}=-1\).

Tangents and Normals at a Point

Differentiation

International Baccalaureate Mathematics: Analysis and Approaches Higher Level

Line	Gradient
Tangent	\(m_{\mathrm{t}}=\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
Normal	\(m_{\mathrm{n}}=-\dfrac{1}{m_{\mathrm{t}}}=-\dfrac{1}{\mathrm{d}y/\mathrm{d}x}\)