MATH 1106, Spring 2017
Tutorial 4 Solutions
Friday 02/03/17
DERIVATIVES 2 SOLUTIONS
Tutorial 4.1. Tangent lines
(a) Find the slope of the tangent line to the graph of the function
f (x) = 2x5 − 3x4 + x2 − 7 at the point x = 1. What is the equation
of the tangent line to the graph of f at this point?
(b) (Ch 4.1 Ex 37) For the function f (x) = 2x3 + 9x2 − 60x + 4 find
the values of x where the tangent is horizontal.
Solution 4.1. (a) The tangent line to f at a is given by the equation
y − f (a) = f 0 (a)(x − a). Here f 0 (x) = 10x4 − 12x3 + 2x by the
rules seen in class. Thus f (1) = −7 and f 0 (1) = 0. The slope
of the tangent line at the point 1 is simply the derivative of f at
1, so this slope is 0, the tangent line is horizontal. Therefore the
tangent at 1 is given by the equation y−(−7) = 0, or equivalently,
y = −7.
(b) The tangent line is horizontal when its slope is zero, hence when
the derivative is zero. The derivative is f 0 (x) = 6x2 + 18x − 60.
Using the quadratic formula, the derivative is zero at:
√
−18 − 182 + 4 · 6 · 60
= −5
x1 =
2·6
and
√
−18 + 182 + 4 · 6 · 60
x2 =
=2
2·6
Tutorial 4.2. Computing derivatives
Using the rules for derivatives seen in class, determine the derivatives of the following functions. If there are values of x for which
you cannot compute the derivative, list these values.
(a) f (x) = x2 + 3x
√+ 5 3
√
(b) f (x) = x + x3 + x
1+x−3x2
(c) f (x) = (x−2)(x+1)
√
(d) f (x) = (x3 + 4x2 − x + 7)(5x5 + x − x1 )
Solution 4.2. (a) f 0√
(x) = 2x + 3
(b) f 0 (x) = 2√1 x + 32 x − x32 when x > 0.
1
2
DERIVATIVES 2 SOLUTIONS
(c)
(1 − 6x)(x − 2)(x + 1) − (1 + x − 3x2 )((x + 1) + (x − 2))
(x − 2)2 (x + 1)2
−3x4 − 4x3 + 13x2 + 14x
=
(x − 2)2 (x + 1)2
−3x2 − 7x
=
(x − 2)(x + 1)
√
(d) f 0 (x) = (3x2 +8x−1)(5x5 + x− x1 )+(x3 +4x2 −x+7)(25x4 + 2√1 x + x12
f 0 (x) =
Tutorial 4.3. The chain rule
Let f and g be differentiable functions (for all real numbers) and let
h = g ◦ f , i.e. for all x, h(x) = g(f (x)). Recall the chain rule: the
derivative of h is given by: h0 (x) = g 0 (f (x)) · f 0 (x).
(a) Describe what the function h(x) does.
(b) Describe what the function h0 (x) does.
(c) Explain the meaning of the chain rule (think of this in the rate of
change interpretation).
Solution 4.3. (a) The function sends each point x first to f (x) then
evaluates g at f (x), namely gives g(f (x)). It applies first the function f , then applies the function g to the resulting value. Think
of this as “every place we see x in the function g(x), replace it by
f (x)”.
(b) The function h0 (x) applies the function g 0 to the function f (x),
(evaluates g 0 at f (x)) then multplies this by f 0 (x). WARNING:
g 0 (f (x)) does not mean we multiply g 0 and f !
(c) Thinking of the derivative as the rate of change, we want to see
how h varies with x. Since h is a composition, we first have to
see how much f changes with x (this is f 0 ), then multiply this
change with how much g changes compared to f (x), not just x,
since we are applying g to f (x), not just to x. This is the g 0 (f (x))
bit.
Tutorial 4.4. Computing using the chain rule
Let f (x) = x2 + 1 and g(x) = 3x − 7.
(a) Compute f [g(x)] and g[f (x)].
(b) Using the chain rule, compute the derivatives of f [g(x)] and g[f (x)].
(c) Using the
√ chain rule, determine the derivative of the function
h(t) = 2 9t2 + 8.
(d) Using the chain rule, determine the derivative of the function
h(t) = (1 + 1t )5
DERIVATIVES 2 SOLUTIONS
3
Solution 4.4. (a) f (g(x)) = f (3x − 7) = (3x − 7)2 + 1 = 9x2 − 42x + 50
and g(f (x)) = g(x2 + 1) = 3(x2 + 1) − 7 = 3x2 − 4. Note that these
are not the same function!
(b) Start by computing f 0 (x) = 2x and g 0 (x) = 3. Then
(f (g(x)))0 = f 0 (g(x)) · g 0 (x) = f 0 (3x − 7) · (3) = 2 · 3(3x − 7) = 18x − 42
and
(g(f (x)))0 = g 0 (f (x)) · f 0 (x) = g 0 (x2 + 1) · (2x) = 3 · 2x = 6x
You can check these against the derivatives obtained by differentiating the polynomials
obtained in (a) directly.
√
(c) Let g(t) = 2 t and f (t) = 9t2 + 8. So f 0 (t) = 18t and g 0 (t) = √1t .
Then h(t) = g(f (t)), so by the chain rule:
1
18t
h0 (t) = g 0 (f (t)) · f 0 (t) = √
· (18t) = √
2
9t + 8
9t2 + 8
which is well-defined when 9t2 + 8 > 0.
(d) Let f (t) = 1 + 1t and g(t) = t5 . Then f 0 (t) = −1
, g 0 (t) = 5t4 and
t2
h(t) = g(f (t)) and by the chain rule:
1
−1
h0 (t) = g 0 (f (t)) · f 0 (t) = 5(1 + )4 · 2
t
t