 # Derivatives in Calculus: Questions with Solutions

Questions on derivatives of functions are presented and their detailed solutions discussed. These questions have been designed to help you to deepen the conceptual understanding of derivatives as well as to develop the computational skills of derivatives.

### Question 1

f, g and H are functions such that
H(x) = (f g)(x)
and
f(1) = 36, f '(-2) = 3, f '(1) = 4, g(1) = 9, g '(1) = -1.
Is the slope of the tangent line to the graph of H at x = 1 positive, negative or equal to zero?

Solution to Question 1:

• G is the product of functions f and g hence.
G '(x) = f '(x) g(x) + f(x) g '(x)
• Substitute x by 1 in the above to obtain.
G '(1) = f '(1) g(1) + f(1) g '(1) = (4)(9) + (36)(-1) = 0
• Since the value of the derivative of H at x = 1 is equal to the slope of the tangent line at x = 1, then the slope is also equal to zero and therefore the tangent line is parallel to the x axis.

### Question 2

Function f is given by
f(x) = a x
2 + b x + c
Find a, b and c so that f (0) = 3, f '(1) = 1 and f "(2) = 4, where f ' and f " are the first and second derivatives of function f.

Solution to Question 2:

• We first use f (0) = 3 to write the following equation
f(0) = a (0) 2 + b (0) + c = 3
• Which gives
c = 3
• We next calculate the first and second derivatives.
f '(x) = 2 a x + b
f "(x) = 2 a
• We first use f "(2) = 4 to find a as follows
f "(2) = 2 a = 4
• Solve for a to find
a = 2
• We now use f '(x) = 2 a x + b and f '(1) = 1 to find b
f '(1) = 2 a (1) + b = 1
• Substitute a by 2 and write an equation in b only
2 (2) (1) + b = 1
• Solve for b to obtain
b = -3
• The values of the 3 parameters a, b and c are
a = 2, b = -3 and c = 3.

### Question 3

Let f be the function defined by f(x) = x3 + x and function g defined by g(x) = f -1(x) (inverse function). what is the value of g'(2)?

Solution to Question 3:

• We first apply function f to both sides of g(x) = f -1(x) and write
f(g(x)) = f(f -1(x)) = x
• We now have an equation of the form f(g(x)) = x which we differentiate (both sides), applying the chain rule to the left side, as follows
f '(g) . g'(x) = 1
• We now set x = 2 in the above equation
f '(g(2)) . g '(2) = 1
• We now calculate the first derivative of f as follows
f '(x) = 3 x 2 + 1
• We now need the value of g(2). Note that f(1) = 2. Therefore g(2) = f -1(2) = 1. Hence
f '(1) . g '(2) = 1
• We now calculate f '(1) using f '(x) = 3 x 2 + 1
f '(1) = 4
• hence g '(2) = 1 / 4

### Question 4

Let functions f, g and h be related as follows:
g(x) = f
-1(x), h(x) = (g(x)) 5, f(6) = 10 and f '(6) = 12
Calculate h '(10).

Solution to Question 4:

• We first calculate h '(x).
h '(x) = 5 g '(x) g(x) 4
• Which gives
h '(10) = 5 g '(10) g(10) 4
• g(10) is calculauted as follows
g(10) = f -1(10) = 6
• We now need to calculate g '(10). Apply function f to both sides of the relation g(x) = f -1(x)
f (g(x)) = f(f -1(x)) = x
Which gives f (g(x)) = x
• Differentiate both sides of the equation obtained, using the chain rule to the term on the left.
f '(g) . g '(x) = 1
• We now set x = 10 in the above equation
f '(g(10)) . g '(10) = 1
• g(10) has already been calculated and is equal to 6, hence
f '(6) . g '(10) = 1
• g '(10) is given by
g '(10) = 1 / f '(6) = 1 / 12
• and h '(10) is given by
h '(10) = 5 (1/12) 6 4 = 540

## More References on Calculus

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