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Simplify first, then differentiate

Consider the function f(x)=1x(x2+3x)f(x) = \dfrac{1}{x}(x^2 + 3x), defined for x≠0x \neq 0.

Simplify f(x)=1x(x2+3x)f(x) = \dfrac{1}{x}(x^2+3x) into a polynomial expression, then differentiate the simplified expression. What is f′(x)f'(x)?

Hint

Distribute and cancel the common factor of xx before taking any derivative — you should be left with a simple linear function.

Step-by-step solution

First distribute: f(x)=x2+3xx=x2x+3xx=x+3f(x) = \dfrac{x^2+3x}{x} = \dfrac{x^2}{x} + \dfrac{3x}{x} = x + 3, valid for x≠0x \neq 0.

Differentiating this simplified linear expression gives f′(x)=1f'(x) = 1.

Differentiate using the quotient rule

Now write f(x)=x2+3xxf(x) = \dfrac{x^2+3x}{x} and compute f′(x)f'(x) directly using the quotient rule, without simplifying first. Show that your result agrees with the answer from part (a).

Hint

Let u=x2+3xu = x^2+3x and v=xv = x. Apply (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}, then simplify the resulting expression algebraically.

Step-by-step solution

With u=x2+3xu = x^2+3x, u′=2x+3u' = 2x+3, and v=xv = x, v′=1v' = 1, the quotient rule gives f′(x)=(2x+3)(x)−(x2+3x)(1)x2f'(x) = \dfrac{(2x+3)(x) - (x^2+3x)(1)}{x^2}.

Expanding the numerator: (2x2+3x)−(x2+3x)=x2(2x^2+3x) - (x^2+3x) = x^2. So f′(x)=x2x2=1f'(x) = \dfrac{x^2}{x^2} = 1, which matches the answer from part (a).

Comparing the two methods' results

Hint
Step-by-step solution

The answer is one. After expanding the numerator in the quotient rule computation, everything cancels except x squared over x squared, leaving the constant one. This matches the derivative found by simplifying first in part a, confirming both methods agree.

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