Question

Given u( – 5) = -5, u'( – 5) = 2, U( – 5) = 6, v'( – 5) = 4, find w'( – 5).

Answer 1

(a) w(x) = 5u(x) + 8v(x)

Differentiating with the sum rule gives

w'(x) = 5u'(x) + 8v'(x)

so that

w' (-5) = 5u' (-5) + 8v' (-5)

… = 5×2 + 8×4 = 42

(b) w(x) = u(x) v(x)

Differentiate using the product rule:

w'(x) = u'(x) v(x) + u(x) v'(x)

Then

w' (-5) = u' (-5) v (-5) + u (-5) v' (-5)

… = 2×6 + (-5)×4 = -8

(c) w(x) = u(x) / v(x)

Quotient rule:

w'(x) = (u'(x) v(x) - u(x) v'(x) ) / v(x) ²

Then

w' (-5) = (u' (-5) v (-5) - u (-5) v' (-5) ) / v (-5)²

… = (2×6 - (-5)×4) / 6² = 32/36 = 8/9

(d) w(x) = u(x) / (u(x) + v(x) )

Chain and quotient rule:

w'(x) = (u'(x) (u(x) + v(x)) - u(x) (u(x) + v(x))' ) / (u(x) + v(x) )²

… = (u'(x) (u(x) + v(x)) - u(x) (u'(x) + v'(x))) / (u(x) + v(x) )²

Then

w' (-5) = (u' (-5) (u (-5) + v (-5)) - u (-5) (u' (-5) + v' (-5))) / (u (-5) + v (-5) )²

… = (2×((-5) + 6) - (-5)×(2 + 4)) / ((-5) + 6)²

… = (2×1 + 5×6) / 1² = 32