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  1. AP Calculus
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Define tan⁡x\tan xtanx.

tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}tanx=cosxsinx​

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Define tan⁡x\tan xtanx.

tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}tanx=cosxsinx​

Define cot⁡x\cot xcotx.

cot⁡x=cos⁡xsin⁡x=1tan⁡x\cot x = \frac{\cos x}{\sin x} = \frac{1}{\tan x}cotx=sinxcosx​=tanx1​

Define sec⁡x\sec xsecx.

sec⁡x=1cos⁡x\sec x = \frac{1}{\cos x}secx=cosx1​

Define csc⁡x\csc xcscx.

csc⁡x=1sin⁡x\csc x = \frac{1}{\sin x}cscx=sinx1​

What is the derivative?

The instantaneous rate of change of a function.

What is the chain rule?

A formula for finding the derivative of a composite function.

What is a composite function?

A function that is composed of another function. For example, f(g(x))f(g(x))f(g(x)).

What is the product rule?

A formula used to differentiate products of two or more functions.

What is the quotient rule?

A method of finding the derivative of a function that is the ratio of two other functions.

What are trigonometric functions?

Functions that relate the angles of a triangle to the lengths of its sides.

Why are the derivatives of co-functions negative?

The derivatives of co-functions (cot, csc) are negative because their original functions are decreasing over their primary intervals.

Explain the importance of the chain rule.

The chain rule is crucial for finding derivatives of composite functions, where one function is nested inside another. It ensures that we account for the rate of change of both the outer and inner functions.

When should you simplify trigonometric expressions before differentiating?

Simplifying trig expressions before differentiating can make the process easier by reducing complexity. Use trig identities to simplify before applying derivative rules.

How does the unit circle relate to derivatives of trigonometric functions?

The unit circle provides a visual representation of trigonometric functions, helping to understand their behavior and derivatives. It shows how sine, cosine, and their related functions change as the angle varies.

Explain the relationship between tan⁡x\tan xtanx and sec⁡x\sec xsecx in differentiation.

The derivative of tan⁡x\tan xtanx is sec⁡2x\sec^2 xsec2x. This means that the rate of change of the tangent function is related to the square of the secant function.

Explain the relationship between cot⁡x\cot xcotx and csc⁡x\csc xcscx in differentiation.

The derivative of cot⁡x\cot xcotx is −csc⁡2x-\csc^2 x−csc2x. This means that the rate of change of the cotangent function is related to the negative of the square of the cosecant function.

Why is it important to use radians when differentiating trigonometric functions?

Using radians ensures that the standard derivative formulas for trigonometric functions are valid. If degrees are used, a conversion factor is needed, making the calculations more complex.

What is the significance of the sign of the derivative?

The sign of the derivative indicates whether a function is increasing (positive derivative) or decreasing (negative derivative).

Explain how derivatives are used in optimization problems.

Derivatives are used to find critical points (where the derivative is zero or undefined) of a function, which can then be used to determine the maximum or minimum values of the function.

What is the relationship between a function and its derivative?

The derivative of a function gives the slope of the tangent line to the function at any given point. It represents the instantaneous rate of change of the function.

What is the derivative of tan⁡x\tan xtanx?

(tan⁡x)′=sec⁡2x(\tan x)' = \sec^2 x(tanx)′=sec2x

What is the derivative of cot⁡x\cot xcotx?

(cot⁡x)′=−csc⁡2x(\cot x)' = -\csc^2 x(cotx)′=−csc2x

What is the derivative of sec⁡x\sec xsecx?

(sec⁡x)′=sec⁡xtan⁡x(\sec x)' = \sec x \tan x(secx)′=secxtanx

What is the derivative of csc⁡x\csc xcscx?

(csc⁡x)′=−csc⁡xcot⁡x(\csc x)' = -\csc x \cot x(cscx)′=−cscxcotx

State the chain rule.

ddxf(g(x))=f′(g(x))cdotg′(x)\frac{d}{dx} f(g(x)) = f'(g(x)) cdot g'(x)dxd​f(g(x))=f′(g(x))cdotg′(x)

State the product rule.

ddx[u(x)v(x)]=u′(x)v(x)+u(x)v′(x)\frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)dxd​[u(x)v(x)]=u′(x)v(x)+u(x)v′(x)

State the quotient rule.

ddx[u(x)v(x)]=u′(x)v(x)−u(x)v′(x)[v(x)]2\frac{d}{dx} \left[ \frac{u(x)}{v(x)} \right] = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}dxd​[v(x)u(x)​]=[v(x)]2u′(x)v(x)−u(x)v′(x)​

What is the derivative of xnx^nxn?

ddxxn=nxn−1\frac{d}{dx} x^n = nx^{n-1}dxd​xn=nxn−1

What is the derivative of a constant ccc?

ddxc=0\frac{d}{dx} c = 0dxd​c=0

What is the derivative of cf(x)cf(x)cf(x)?

ddxcf(x)=cddxf(x)\frac{d}{dx} cf(x) = c \frac{d}{dx} f(x)dxd​cf(x)=cdxd​f(x)