### Derivative : solution exercise 1 (Intermediate math level)

Given the quadratic equation y = 3x2 + 2 and the function y = (1/2)x3 - 1, you are asked to :

a) Calculate the derivative y ' of both functions

• y ' = (3x2 + 2) ' = 6x

• y ' = (0,5x3 - 1) ' = 1,5x2 = (3/2)x2

b) Compute, for both curves, the slope of their tangent lines at x = 1 using the derivatives figured out hereabove

The slope of the tangent line to a curve at x = 1 is the value of its derivative at x = 1.

• y ' = f '(x) = 6x
f '(1) = 6 . 1 = 6 = slope of the tangent line to the curve y = 3x2 + 2 at x = 1

• y ' = f '(x) = (3/2)x2
f '(1) = (3/2) . 12 = 3/2 = slope of the tangent line to the curve y = (1/2)x3 - 1
at x = 1

c) Find the equation of both tangent lines

• For y = 3x2 + 2, its tangent line at x = 1 is the tangent line to the curve at the point
x = 1 and y = 3 . 12 + 2 = 5, thus I know that my tangent line passes through the point (1 ; 5). I also know that the tangent line has a slope of a = 6. Let us now compute the equation of this tangent line :

General cartesian equation of a straight line : y = ax + b

Equation of the tangent line : y = 6x + b ...

... that passes through the point (1 ; 5) : 5 = 6 . 1 + b

thus 5 - 6 = b

Which yields b = -1

Conclusion : the tangent line to the curve y = 3x2 + 2 at x = 1 is y = 6x - 1

• We use the same method to figure out the equation of the tangent line to the curve
y = (1/2)x3 - 1 at x = 1

The tangent line passes through the point that has an x-coordinate = 1
and an y-coordinate = 0,5 . 13 - 1 = -0,5 = -1/2

Equation of the tangent line that has a slope a = 3/2 through the point (1 ; -1/2) :

y = (3/2)x + b

-1/2 = (3/2) . 1 + b

Thus -(1/2) - (3/2) = b

Which yields b = -4/2 = -2

Conclusion : the tangent line to the curve y = (1/2)x3 - 1 at x = 1 is
y = (3/2)x - 2

d) Trace both curves and its tangent lines on a graph

Both graphs have been drawn with our free online Graph Plotter which is a real scientific graphing calculator that enables you to trace any mathematical function online.