APPLICATIONS OF
DIFFERENTIATION
SOME APPLICATIONS
,
,F’(x) measures the gradient of the tangent to the curve y=f(x) at the point w
abscissa x. The gradient of the tangent at the point with abscissa x1 is then
Thus the equation of the tangent to the curve at the point (x1, y1) on the cu
by (18.6)
𝑦 − 𝑦1 = 𝑓 ′ 𝑥1 𝑥 − 𝑥1
The normal to the curve at the point at the point (x1, y1) on the curve is the
through this point perpendicular to the tangent to the curve. The gradient
normal is thus -1/f’(x1) and the equation of the normal is, by,
−1
𝑦 − 𝑦1 = ′ 𝑥 − 𝑥1
𝑓 𝑥1
, Example 1 Find the equation of the tangent and normal to the curve y=3x2 – 5x at the poin
2)
𝑦 = 𝑓 𝑥 = 3𝑥 2 − 5𝑥
𝑓 ′ 𝑥 = 6𝑥 − 5
𝑓′ 1 = 6 ∙ 1 − 5 = 1
Therefore, the tangent at (1,-2) has the equation
𝑦 − −2 = +1 𝑥 − 1
𝑦+2=𝑥−1
𝑦 =𝑥−3
The normal at (1, -2) has the equation
𝑦 − −2 = −1 𝑥 − 1
𝑦 = −𝑥 − 1
DIFFERENTIATION
SOME APPLICATIONS
,
,F’(x) measures the gradient of the tangent to the curve y=f(x) at the point w
abscissa x. The gradient of the tangent at the point with abscissa x1 is then
Thus the equation of the tangent to the curve at the point (x1, y1) on the cu
by (18.6)
𝑦 − 𝑦1 = 𝑓 ′ 𝑥1 𝑥 − 𝑥1
The normal to the curve at the point at the point (x1, y1) on the curve is the
through this point perpendicular to the tangent to the curve. The gradient
normal is thus -1/f’(x1) and the equation of the normal is, by,
−1
𝑦 − 𝑦1 = ′ 𝑥 − 𝑥1
𝑓 𝑥1
, Example 1 Find the equation of the tangent and normal to the curve y=3x2 – 5x at the poin
2)
𝑦 = 𝑓 𝑥 = 3𝑥 2 − 5𝑥
𝑓 ′ 𝑥 = 6𝑥 − 5
𝑓′ 1 = 6 ∙ 1 − 5 = 1
Therefore, the tangent at (1,-2) has the equation
𝑦 − −2 = +1 𝑥 − 1
𝑦+2=𝑥−1
𝑦 =𝑥−3
The normal at (1, -2) has the equation
𝑦 − −2 = −1 𝑥 − 1
𝑦 = −𝑥 − 1
