Find an equation of the straight line having slope 1/4 that is tangent to the curve y = √x.
Ch. 3 - Derivatives
Chapter 3, Problem 3.6.78
Find dy/dt when x = 1 if y = x² + 7x − 5 and dx/dt = ¹/₃.
Verified step by step guidance1
First, identify the given function y = x² + 7x − 5. This is the function for which we need to find the derivative with respect to time, t.
To find dy/dt, use the chain rule. The chain rule states that dy/dt = (dy/dx) * (dx/dt). We need to find dy/dx first.
Differentiate y = x² + 7x − 5 with respect to x to find dy/dx. The derivative of x² is 2x, and the derivative of 7x is 7. Therefore, dy/dx = 2x + 7.
Substitute x = 1 into the expression for dy/dx to find the specific value of the derivative at x = 1. This gives dy/dx = 2(1) + 7.
Now, substitute dy/dx and dx/dt = ¹/₃ into the chain rule formula: dy/dt = (dy/dx) * (dx/dt). Calculate dy/dt using the values obtained in the previous steps.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Implicit Differentiation
Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. In this context, it helps us differentiate y with respect to t by considering the relationship between x and y, and using the chain rule to account for dx/dt.
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Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental concept in calculus used to differentiate composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. Here, it allows us to find dy/dt by multiplying dy/dx by dx/dt.
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Intro to the Chain Rule
Substitution
Substitution involves replacing variables with known values to simplify expressions or solve equations. In this problem, after finding the expression for dy/dt, we substitute x = 1 and dx/dt = 1/3 to calculate the specific value of dy/dt at that point, providing a concrete solution to the problem.
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Substitution With an Extra Variable
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