Product And Quotient Rules
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Product And Quotient Rules
We have seen how to compute derivatives of sums, differences and constant multiples of functions. In this section, we find the derivatives of products and quotients. Since the derivative of a sum of functions is the sum of their derivatives, one might think that the derivate of a product of tow functions is the product of their derivatives. However, this is not the case, as the next rule shown.
Rule (Product Rule). If f and g are differentiable functions, then
d/dx [ f (x) g (x) ] = f (x) d/dx [ g (x) ] + g (x) d/dx [f (x)].
That is, the derivative of the product of two functions is the first function times the derivative of the derivative of the second, plus the second functions times the derivative of the first.
Proof. Let F (x) = f (x) g (x), then by the definition of the derivative, we have
F ' (x) = lim F (x+h) - F (x)
h→0 h
= lim f ( x+h)g (x+h)= f (x)g(x)
h→0 h
= lim f (x+h)g(x+h) - f (x)g(x) + [f (x+h) g (x) - f (x)g (x)]
h→0 h
(Adding and subtracting f (x + h) g (x) in the numerator)
= lim f (x+h) g (x+h) - f (x+h) g (x) + [f (x+h) g (x) - f (x) g (x)]
h→0 h
lim f (x+h) [g (x+h) - g (x)] + g (x) [f (x+h) - f (x)]
h→0 h
= lim f (x+h) [g (x+h) - g (x)] + lim g (x) [f (x+h) - f (x)]
h→0 h h→0 h
= lim f (x+h) . lim g (x+h) - g (x)
h→0 h
+ lim g (x) . lim f (x+h) - f (x)
h→0 h→0 h
Since f is differentiable, it is continuous, and hence
lim f (x+h) = f (x)
h→0
Thus
F ‘ (x) = f (x) g’ (x) + g (x) f ‘ (x).
Rule (Quotient Rules) If f and g are differentiable functions and g (x) ≠ 0, then
d/dx [f (x)/g (x)] = g (x) f ' (x) - f (x)g ' (x)
[g (x)]2
That is, the derivative of the quotient of two functions is the denominator times the derivative of the numerator. minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
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Rule (Product Rule). If f and g are differentiable functions, then
d/dx [ f (x) g (x) ] = f (x) d/dx [ g (x) ] + g (x) d/dx [f (x)].
That is, the derivative of the product of two functions is the first function times the derivative of the derivative of the second, plus the second functions times the derivative of the first.
Proof. Let F (x) = f (x) g (x), then by the definition of the derivative, we have
F ' (x) = lim F (x+h) - F (x)
h→0 h
= lim f ( x+h)g (x+h)= f (x)g(x)
h→0 h
= lim f (x+h)g(x+h) - f (x)g(x) + [f (x+h) g (x) - f (x)g (x)]
h→0 h
(Adding and subtracting f (x + h) g (x) in the numerator)
= lim f (x+h) g (x+h) - f (x+h) g (x) + [f (x+h) g (x) - f (x) g (x)]
h→0 h
lim f (x+h) [g (x+h) - g (x)] + g (x) [f (x+h) - f (x)]
h→0 h
= lim f (x+h) [g (x+h) - g (x)] + lim g (x) [f (x+h) - f (x)]
h→0 h h→0 h
= lim f (x+h) . lim g (x+h) - g (x)
h→0 h
+ lim g (x) . lim f (x+h) - f (x)
h→0 h→0 h
Since f is differentiable, it is continuous, and hence
lim f (x+h) = f (x)
h→0
Thus
F ‘ (x) = f (x) g’ (x) + g (x) f ‘ (x).
Rule (Quotient Rules) If f and g are differentiable functions and g (x) ≠ 0, then
d/dx [f (x)/g (x)] = g (x) f ' (x) - f (x)g ' (x)
[g (x)]2
That is, the derivative of the quotient of two functions is the denominator times the derivative of the numerator. minus the numerator times the derivative of the denominator, all divided by the square of the denominator.
For more help in Product And Quotient Rules click the button below to submit your homework assignment