Slopes of Tangents (Leaving Cert Mathematics): Revision Notes

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Slopes of Tangents

We've seen how to find the slope of a tangent at any given xx point using the first derivative :

Example

infoNote

Find the slope of the tangent at x=12x=\tfrac{1}{2} for the function f(x)=3x2−2x3f(x)=3x^2-2x^3.

Find the first derivative :

f′(x)=6x−6x2f'(x)=6x-6x^2

This represents the slope for any xx, for a particular xx, just substitute the value into xx :

f′(12)=6(12)−6(12)2=32f'(\tfrac{1}{2})=6(\tfrac{1}{2})-6(\tfrac{1}{2})^2=\tfrac{3}{2}

Now find the equation of the line with the slope 32\tfrac{3}{2}.

m=32m=\tfrac{3}{2}

We also need the coordinate :

x=12x=\frac{1}{2}y=3(12)2−2(12)3=12y=3(\tfrac{1}{2})^2-2(\tfrac{1}{2})^3=\tfrac{1}{2}

Now find the equation of the line :

y−y1=m(x−x1)y-y_1=m(x-x_1)y−(12)=32(x−12)y-(\tfrac{1}{2})=\tfrac{3}{2}(x-\tfrac{1}{2})y=32x−14y=\tfrac{3}{2}x-\tfrac{1}{4}

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The Basics

Curve Sketching

Implicit Differentiation

Rates of Change

Trig/Exponential Functions

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