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Chapter 3: Problem 15

a. Use the Product Rule to find the derivative of the given function. Simplify your result. b. Find the derivative by expanding the product first. Verify that your answer agrees with part \((a)\) $$f(x)=(x-1)(3 x+4)$$

Short Answer

Expert verified
Question: Find the derivative of the function $$f(x)=(x-1)(3x+4)$$ using both the Product Rule and by expanding the product and differentiating the resulting polynomial. Verify that both methods give the same result. Answer: Both methods yield the same derivative: $$f'(x) = 6x + 1$$.

Step by step solution

01

Find the derivative using the Product Rule

Recall that the Product Rule states: if $$f(x) = u(x)v(x)$$, then $$f'(x) = u'(x)v(x) + u(x)v'(x)$$. Here, our function is $$f(x) = (x-1)(3x+4)$$, so we can consider $$u(x) = x-1$$ and $$v(x) = 3x + 4$$. First, we need to find the derivatives of $$u(x)$$ and $$v(x)$$: $$u'(x) = \frac{d}{dx}(x-1) = 1$$ $$v'(x) = \frac{d}{dx}(3x+4) = 3$$ Now, we can apply the Product Rule: $$f'(x) = u'(x)v(x) + u(x)v'(x) = (1)(3x+4) + (x-1)(3)$$
02

Simplify the result from Step 1

Now we will simplify the result from Step 1: $$f'(x) = (1)(3x+4) + (x-1)(3) = 3x + 4 + 3x - 3 = 6x + 1$$
03

Find the derivative by expanding the product first

Now we will expand the product of $$f(x)$$ and then find the derivative: $$f(x) = (x-1)(3x+4) = x(3x+4) - 1(3x+4) = 3x^2 + 4x - 3x - 4 = 3x^2 + x - 4$$ Next, we will find the derivative: $$f'(x) = \frac{d}{dx}(3x^2 + x - 4) = 6x + 1$$
04

Compare the results

Finally, we will compare the results from Step 2 and Step 3. We found that both methods yielded the same derivative: $$f'(x) = 6x + 1$$. Therefore, we have successfully verified that both methods agree.

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