Textbook Question
9–61. Evaluate and simplify y'.
y = (2x−3)x^3/2
3. Techniques of Differentiation
Product and Quotient Rules
8:01 minutesProblem 3.4.61a
Textbook Question
Find an equation of the line tangent to the given curve at a.
y = (x + 5) / (x - 1); a = 3
Verified step by step guidance1
Step 1: Identify the function and the point of tangency. The function is \( y = \frac{x + 5}{x - 1} \) and the point of tangency is at \( x = 3 \).
Step 2: Calculate the derivative of the function to find the slope of the tangent line. Use the quotient rule: if \( y = \frac{u}{v} \), then \( y' = \frac{u'v - uv'}{v^2} \). Here, \( u = x + 5 \) and \( v = x - 1 \).
Step 3: Differentiate \( u \) and \( v \). \( u' = 1 \) and \( v' = 1 \). Substitute these into the quotient rule to find \( y' \).
Step 4: Evaluate the derivative at \( x = 3 \) to find the slope of the tangent line. Substitute \( x = 3 \) into the expression for \( y' \).
Step 5: Use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope found in Step 4 and \( (x_1, y_1) \) is the point \( (3, y(3)) \). Calculate \( y(3) \) using the original function.
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
8mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Tangent Line
A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line represents the instantaneous rate of change of the function at that point, which can be found using the derivative.
Recommended video:
05:13
Slopes of Tangent Lines
Derivative
The derivative of a function measures how the function's output value changes as its input value changes. It is defined as the limit of the average rate of change of the function over an interval as the interval approaches zero. For the function y = (x + 5) / (x - 1), the derivative will provide the slope of the tangent line at the point where x = 3.
Recommended video:
05:44Derivatives
Point-Slope Form
The point-slope form of a linear equation is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is the slope. This form is particularly useful for writing the equation of a tangent line once the slope (from the derivative) and the point of tangency (the coordinates of the curve at x = 3) are known.
Recommended video:
Guided course
3:56Slope-Intercept Form
Related Videos
Related Practice