Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (e)/Solution 1

To find the equation of the tangent line to y = x^3.5 - e^3.5 at the point (e, 0 ), we must first find the derivative of the curve, which is the slope of the tangent line at that point.

First we use the difference rule to separate the two terms.

${\frac {d}{dx}}(x^{3.5}-e^{3.5})={\frac {d}{dx}}(x^{3.5})-{\frac {d}{dx}}(e^{3.5})$

The next step is to recognize that e^3.5 is a constant, so we know that its derivative is equal to 0 and we can remove that term altogether from the equation. Then we use to power rule to differentiate x^3.5 :

${\frac {d}{dx}}(x^{3.5})=3.5x^{3.5-1}=3.5x^{2.5}$

By plugging in x = e into the derivative, we find the tangent line at the desired point (e, 0), we get the slope m of the tangent line of equation y-y₀ = m(x - x₀)

$\displaystyle m=3.5(e^{2.5})$

From this we can conclude that the equation of the line is:

$\displaystyle y-y_{0}=m(x-x_{0})\quad \Rightarrow \quad y-0=3.5e^{2.5}(x-e)$

Simplifying, we get

$\displaystyle y=3.5e^{2.5}(x-e)$

or

$\displaystyle y=3.5e^{2.5}x-3.5e^{3.5}$