Differentiation rules practice question 3
Find an equation of the tangent line to at the point where .
A line is fixed by a point and a slope. Which two facts about the curve at supply them?
The point is and the slope is , where . Then use point-slope form.
An equation of a line. Use ^ for powers and brackets for every input: sin(2x), e^(x^2), log(x+1). Give exact values: 1/6, not 0.167.
Theory UBC Math 100 §4.1CLP-1 §2.4Key results
Worked solution
Strategy
Find the point of tangency and the slope there, then write the line through that point with that slope.
Working
The point of tangency is (2, 14).
Power rule term by term, then substitute.
Point-slope form, then simplify.
Answer
The answer is y equals 23x minus 32.
Check
At x = 2 the line gives 14, the height of the curve, and at x = 2.01 the curve and the line are both about 14.23.
Where this goes wrong
Writing y = 23x + 14, with y(2) = 14 as the intercept. The line passes through (2, 14), so its value at x = 0 is 14 − 2 · 23 = −32.
The idea behind it
The differentiation rules lesson, a figure to move, in a few steps
Leans on
Point-slope form of a line, from Functions, graphs and inverses
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