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Maths

Differentiation

Tangents to curves

It is very straightforward to obtain the equation of the tangent to a curve with equation y = f(x) provided you can differentiate f(x).

Follow this worked example which shows you how to find the equation of the tangent to the curve with equation y = {1 \over 8}x^3 - 3\sqrt x at the point where {x =4 }.

First, you have to find the y-coordinate of the point where {x =4 }

y = {1 \over 8}4^3 - 3\sqrt 4 = 8 - 6 = 2 \Rightarrow (4,2)

Now differentiate y'(x) = {3 \over 8}x^2 - {3 \over 2}x^{ - {1 \over 2}}

(knowing to differentiate could be worth a mark)

The gradient at (4, 2) is the value of y' when {x =4 }; in other words, y'(4)

y'(4) = {3 \over 8}4^2 - {3 \over 2}4^{ - {1 \over 2}} = 6 - {3 \over 4} = {{21} \over 4}

Now you need to find the equation of the line through (4, 2) with gradient {{21} \over 4}.

To do this, you use the >y - b = m (x - a) formula from the straight line section, often also written as y - y_1 = m\;(x - x_1 )

\eqalign{ y - 2 & = & {{21} \over 4}(x - 4)\cr\cr \Rightarrow 4y - 8 & = & 21x - 84\cr\cr \Rightarrow 4y & = & 21x - 76 \cr}

Question

Find the equation of the tangent to the curve with equationy = 3x^2 + 2x - 5 at the point where x = -2.

Answer

y( - 2) = 3 \times ( - 2)^2 + 2 \times ( - 2) - 5 = 12 - 4 - 5 = 3
so the point is (-2,3).

y' \left( x \right) =\;\;6x + 2 \Rightarrow gradient at (-2, 3) = y'(-2) = 6(-2) + 2 = -10

hence y - 3 = -10 (x - (-2)) => 10x + y + 17 = 0

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