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3 The Derivative and Differentiation Rules Free Online FlashCards

Study 3 The Derivative and Differentiation Rules with 12 free online flashcards. Review key terms, definitions, and concepts with this interactive flashcard deck.

12 cards
01
Front

How is f′(a)f'(a) defined by a limit?

Back

The derivative is the limit
inline math
f′(a)=lim⁡h→0f(a+h)−f(a)hf'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}
when the limit exists. It measures instantaneous change at aa.

02
Front

What does f′(a)f'(a) represent geometrically?

Back

At (a,f(a))(a,f(a)), f′(a)f'(a) is the slope of the tangent line. The tangent-line equation is y−f(a)=f′(a)(x−a)y-f(a)=f'(a)(x-a).

03
Front

Does differentiability imply continuity?

Back

Differentiability at a point implies continuity there, but continuity does not always imply differentiability. For example, f(x)=∣x∣f(x)=|x| is continuous but not differentiable at x=0x=0.

04
Front

What is the power rule?

Back

The power rule is ddx[xn]=nxn−1\frac{d}{dx}[x^n]=nx^{n-1}, where the expression is defined.

05
Front

What is the product rule?

Back

The product rule is ddx[f(x)g(x)]=f′(x)g(x)+f(x)g′(x)\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x).

06
Front

What is the quotient rule?

Back

The quotient rule is ddx[f(x)g(x)]=g(x)f′(x)−f(x)g′(x)[g(x)]2\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}, for g(x)≠0g(x)\ne 0.

07
Front

What is the derivative of exe^x?

Back

The derivative of exe^x is exe^x itself: ddx[ex]=ex\frac{d}{dx}[e^x]=e^x.

08
Front

What is the derivative of log⁡ax\log_a x?

Back

For a>0a>0 and a≠1a\ne 1, ddx[log⁡ax]=1xln⁡a\frac{d}{dx}[\log_a x]=\frac{1}{x\ln a}.

09
Front

How does the chain rule differentiate f(g(x))f(g(x))?

Back

The chain rule gives ddxf(g(x))=f′(g(x))g′(x)\frac{d}{dx}f(g(x))=f'(g(x))g'(x): differentiate the outer function, keep the inner expression, then multiply by the inner derivative.

10
Front

Differentiate (3x2−5)4(3x^2-5)^4.

Back

Using the chain rule, ddx[(3x2−5)4]=4(3x2−5)3(6x)=24x(3x2−5)3\frac{d}{dx}[(3x^2-5)^4]=4(3x^2-5)^3(6x)=24x(3x^2-5)^3.

11
Front

How is yny^n differentiated implicitly?

Back

When differentiating with respect to xx, treat yy as a function of xx: ddx[yn]=nyn−1dydx\frac{d}{dx}[y^n]=ny^{n-1}\frac{dy}{dx}.

12
Front

Find dydx\frac{dy}{dx} for x2+y2=25x^2+y^2=25.

Back

Differentiating x2+y2=25x^2+y^2=25 gives 2x+2ydydx=02x+2y\frac{dy}{dx}=0, so dydx=−xy\frac{dy}{dx}=-\frac{x}{y}.