Question

Derivatives from tangent lines Suppose the line tangent to the graph of \(f\) at \(x=2\) is \(y=4 x+1\) and suppose \(y=3 x-2\) is the line tangent to the graph of \(g\) at \(x=2 .\) Find an equation of the line tangent to the following curves at \(x=2.\) a. \(y=f(x) g(x)\) b. \(y=f(x) / g(x)\)

Short answer

In this problem, we find the tangent lines to the curves y = f(x)g(x) and y = f(x)/g(x) at x = 2. The tangent lines to f(x) and g(x) at x = 2 are given as y = 4x + 1 and y = 3x - 2. We find the derivatives f'(2) = 4 and g'(2) = 3, then apply the product and quotient rules to find the derivatives of the combined functions at x = 2, resulting in a slope of 43 for f(x)g(x) and -3/4 for f(x)/g(x). The equations we derive for the tangent lines at x = 2 are as follows: a. For y = f(x)g(x): y = 43x - 50 b. For y = f(x)/g(x): y = -3/4 * x + 15/4

Step-by-step solution

Find the derivatives of f(x) and g(x) at x = 2#

To find the derivatives of f(x) and g(x) at x = 2, we need their tangent lines equations given as: \(y = 4x + 1\) for \(f(x)\) and \(y = 3x - 2\) for \(g(x)\). The slope of each tangent line will give us the derivative at that point. For f(x), we have: slope, \(f'(2) = 4\). For g(x), we have: slope, \(g'(2) = 3\). So we find that \(f'(2) = 4\) and \(g'(2) = 3\).

Find the derivative of y = f(x)g(x) at x = 2#

Using the product rule, the derivative \(h'(x)\) of \(h(x)=f(x)g(x)\) is given by: \(h'(x) = f'(x)g(x) + f(x)g'(x)\) Now to find \(h'(2)\), we need to find \(f(2)\) and \(g(2)\) using the tangent lines equations. For f(x), at x = 2: \(f(2)= 4(2) +1= 9\). For g(x), at x = 2: \(g(2) = 3(2) -2 = 4\). Now we find \(h'(2)\) as follows: \(h'(2) = f'(2)g(2) + f(2)g'(2) = (4)(4) + (9)(3) = 43\)

Find the equation of the tangent line for y = f(x)g(x) at x = 2#

For this, we use the point-slope form of the tangent line equation with the slope \(h'(2) = 43\) and the point \((2, f(2)g(2)) = (2, 9\cdot4= 36)\). Therefore, the tangent line is given by: \(y - 36 = 43(x - 2)\) So the equation of the tangent line is: \(y = 43x - 50\).

Find the derivative of y = f(x)/g(x) at x = 2#

Using the quotient rule, the derivative \(k'(x)\) of \(k(x)=\frac{f(x)}{g(x)}\) is given by: \(k'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}\) Now, we find \(k'(2)\) as follows: \(k'(2)= \frac{f'(2)g(2) - f(2)g'(2)}{g^2(2)} = \frac{(4)(4) - (9)(3)}{4^2}=-\frac{3}{4}\)

Find the equation of the tangent line for y = f(x)/g(x) at x = 2#

For this, we use the point-slope form of the tangent line equation with the slope \(k'(2) = -\frac{3}{4}\) and the point \((2, \frac{f(2)}{g(2)}) = (2, \frac{9}{4})\). Therefore, the tangent line is given by: \(y - \frac{9}{4} = -\frac{3}{4}(x - 2)\) So the equation of the tangent line is: \(y = -\frac{3}{4}x + \frac{15}{4}\). The equations of the tangent lines for the given curves at \(x = 2\) are as follows: a. \(y = f(x)g(x)\): \(y = 43x - 50\) b. \(y = \frac{f(x)}{g(x)}\): \(y = -\frac{3}{4}x + \frac{15}{4}\)

Key concepts

Product Rule

The product rule is a fundamental technique in calculus used to find the derivative of a function that is the product of two other functions. If you have two functions, say f(x) and g(x), and you need to find the derivative of their product, h(x) = f(x)g(x), the product rule tells you that the derivative, h'(x), is found by multiplying the derivative of the first function by the second function and adding the product of the first function and the derivative of the second function:

Product Rule Formula:

\[h'(x) = f'(x)g(x) + f(x)g'(x)\]
This rule is incredibly useful when dealing with complex functions where direct multiplication and subsequent differentiation would be cumbersome. For example, if the functions are f(x) = x^2 and g(x) = sin(x), using the product rule simplifies the process of finding the derivative.

Quotient Rule

Similar to the product rule, the quotient rule is a derivative technique used for functions divided by each other. When you have a function in the form of k(x) = f(x)/g(x), and you want to differentiate it, the quotient rule is what you apply. The rule states that the derivative of such a function, k'(x), can be found by subtracting the product of the derivative of the numerator and the denominator from the product of the numerator and the derivative of the denominator, all over the square of the denominator function:

Quotient Rule Formula:

\[k'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g^2(x)}\]
This rule is particularly important to remember because directly differentiating the numerator and the denominator separately before dividing them would not give the correct result. For example, if f(x) = x^3 and g(x) = e^x, the quotient rule would enable you to efficiently find the derivative of the division of these two functions.

Derivatives

Derivatives are one of the most important concepts in calculus. They measure how a function changes as its input changes; essentially, a derivative represents the slope of the function at any given point. In practical terms, the derivative gives us the rate of change or the instantaneous velocity of an object if we're talking about a position-time graph.

Finding the derivative, or differentiating a function, involves applying rules like the aforementioned product and quotient rules. Other common rules include the power rule, chain rule, and derivative of elementary functions like exponential and trigonometric functions. Calculating the derivative at a particular input value, such as x = 2, gives you the slope of the tangent line to the function's graph at that point, which is critical in many applications about motion, optimization, and curve sketching.

Point-Slope Form

Once you've found the derivative of a function at a specific point, and therefore the slope of the tangent line at that point, you can use the point-slope form to write the equation of the tangent line. The point-slope form is given by:

Point-Slope Form Equation:

\[y - y_1 = m(x - x_1)\]
where m is the slope of the line and (x_1, y_1) is a point on the line. This form is used directly in calculus to find tangent lines to curves at certain points. It's a straightforward method—once you have the slope from the derivative and the coordinates of the point of tangency, you are able to construct the tangent line equation efficiently. This is extremely practical for visualizing the behavior of functions and for solving real-world problems where the tangent line represents an important concept, such as the best linear approximation of a function at a given point.