AP Calculus BC Unit 3 Differentiation-Composite, Implicit, and Inverse Functions

Unit 3: Differentiation - Composite, Implicit, and Inverse Functions ● Higher Order Derivatives: ○ If you get a fraction for a function, or a square root: ■ Ex. Find the derivative of 3/x ● Rewrite the function so that you can apply the power rule to it ○ 3/x = 3𝑥 −1 , and now it’s easier to use the power rule ■ Ex. Find the derivative of √x (square root of x) ● Rewrite → √x = 𝑥 1/2 , and then use the power rule to find the derivative ● Chain Rule: ○ Used to find the derivative of composite functions ■ Composite functions : f(g(x)) ○ Chain rule formula: ■ Ex. y= 4(𝑥 2 − 3𝑥)2 ● f(x) = 4𝑥 2 ; g(x) = 𝑥 2 -3x ● Using the formula, you’d get 8 (𝑥 2 - 3x)(2x-3), which is the answer ● Graphing derivatives: ○ When given a graph of f(x), any point where the tangent line at that point is horizontal, means that the derivative at that point is 0 (because slope = 0) ■ These typically occur at local maximums or minimums of the graph, and they are the zeroes of a f’(x) graph ○ To determine if the derivative of a function is negative or positive at a certain point or interval, just look at the slope of the graph. Whenever the slope is negative, that means the derivative is negative, and if it’s positive, then the derivative is positive. ○ The second derivative measures concavity of the function. If you’re given the graph of f(x), and the graph is concave up, meaning it has a ‘u’ shape to it, then the SECOND derivative is positive. If the graph is concave down, with an upside down ‘u’ shape, then the SECOND derivative is negative. ○ To find the equation of a tangent line when given an equation, just take the derivative of the original function and plug in the given x value to determine the slope of the tangent line equation. If the y value wasn’t already given for that specific x value, determine the y value by plugging the x value into the original equation. Then using the slope formula y = mx + b, you can figure out the equation of the tangent line. ● Implicit Differentiation: ○ When you can’t isolate the function ‘y’ in terms of x, you need to use implicit differentiation. ■ Ex. Find dy/dx for the equation 𝑦 2 -3y+2x = 3 ● To solve this problem, you tackle it the way you’d normally do, except whenever you take the derivative of any ‘y’ functions, you have to denote dy/dx too (look at next bullet point) ● 2y(dy/dx) - 3(dy/dx) + 2 (dx/dx) = 0 ○ *Remember, when taking the derivative of y in respect to x, you get dy/dx. Likewise, taking the derivative of x in respect to x will get you dx/dx, which is 1, and thus, it’s usually implied and not written out. ○ Once you’ve taken the derivative of the whole equation, you just need to solve for dy/dx by isolating dy/dx ■ (dy/dx)(2y - 3) = -2 → dy/dx = -2/(2y - 3) ■ If the problem asks you to find the derivative of y in respect to t, you do the same thing, except for now it’s dy/dt and dx/dt, and you now have to write dx/dt after you take the derivative of the x function too ● Ex. 2y - 3x → derivative would be 2(dy/dt) - 3 (dx/dt) ○ To find the second derivative, take the derivative of the first derivative (after you isolated (dy/dx). Make sure you also apply the rules of implicit differentiation when you take the second derivative. Then, whenever plug in the original equation of dy/dx into your new second derivative to get your answer. ■ Using the example from above, dy/dx = -2/(2y - 3) ■ You need to use the quotient rule to find the second derivative of this equation → f’’(x) = [(2y-3)(0)-(-2)(2(dy/dx))] / (2𝑦 − 3)2 → 4(dy/dx) / (2𝑦 − 3)2 ● Next, plug in the original equation of dy/dx and simplify ○ f’’(x) = 4(-2/(2y-3)) / (2𝑦 − 3)2→ f’’(x) = (-8/(2y3))/ (2𝑦 − 3)2 → f’’(x) = -8/ (2𝑦 − 3)3 ● Derivative of an inverse function: ○ *Remember: to find the equation of the inverse function, you can plug in the given ‘x’ value into ‘y’ instead of ‘x’ as you’d normally do