All flashcards Flashcard 1: Identify u u u in h ( x ) = ( 3 x 2 + 2 x ) 5 h(x) = (3x^2 + 2x)^5 h ( x ) = ( 3 x 2 + 2 x ) 5 . Answer: u = 3 x 2 + 2 x u = 3x^2 + 2x u = 3 x 2 + 2 x . The expression being raised to the 5th power.
Flashcard 2: Find the derivative of sin 2 ( x ) \text{sin}^2(x) sin 2 ( x ) . Answer: 2 sin ( x ) cos ( x ) 2\text{sin}(x)\text{cos}(x) 2 sin ( x ) cos ( x ) . Chain rule: 2 sin ( x ) × cos ( x ) 2\sin(x) \times \cos(x) 2 sin ( x ) × cos ( x ) or sin ( 2 x ) \sin(2x) sin ( 2 x ) .
Flashcard 3: Differentiate y = sin ( ln ( x 2 ) ) y = \text{sin}(\text{ln}(x^2)) y = sin ( ln ( x 2 )) . Answer: 2 x cos ( ln ( x 2 ) ) x 2 \frac{2x\text{cos}(\text{ln}(x^2))}{x^2} x 2 2 x cos ( ln ( x 2 )) . Chain rule twice: cos ( ln ( x 2 ) ) × 2 x x 2 \cos(\ln(x^2)) \times \frac{2x}{x^2} cos ( ln ( x 2 )) × x 2 2 x .
Flashcard 4: Differentiate f ( x ) = ln ( e x + 1 ) f(x) = \text{ln}(\text{e}^x + 1) f ( x ) = ln ( e x + 1 ) . Answer: e x e x + 1 \frac{\text{e}^x}{\text{e}^x + 1} e x + 1 e x . Chain rule: 1 u × u ′ \frac{1}{u} \times u' u 1 × u ′ where u = e x + 1 u = e^x + 1 u = e x + 1 .
Flashcard 5: Find d y / d x dy/dx d y / d x if y = ln ( e x 2 ) y = \text{ln}(\text{e}^{x^2}) y = ln ( e x 2 ) . Answer: 2 x 2x 2 x . Simplifies to x 2 x^2 x 2 since ln \ln ln and e e e cancel.
Flashcard 6: Identify u u u in y = ( ln ( x ) ) 4 y = (\text{ln}(x))^4 y = ( ln ( x ) ) 4 . Answer: u = ln ( x ) u = \text{ln}(x) u = ln ( x ) . The natural logarithm function raised to power 4.
Flashcard 7: Differentiate y = e tan ( x ) y = \text{e}^{\text{tan}(x)} y = e tan ( x ) . Answer: sec 2 ( x ) e tan ( x ) \text{sec}^2(x)\text{e}^{\text{tan}(x)} sec 2 ( x ) e tan ( x ) . Chain rule: e u × u ′ e^u \times u' e u × u ′ where u = tan ( x ) u = \tan(x) u = tan ( x ) .
Flashcard 8: Which function is the inner function in f ( g ( x ) ) = e 3 x + 7 f(g(x)) = \text{e}^{3x+7} f ( g ( x )) = e 3 x + 7 ? Answer: g ( x ) = 3 x + 7 g(x) = 3x+7 g ( x ) = 3 x + 7 . The linear expression in the exponent.
Flashcard 9: What is y ′ y' y ′ if y = cos ( 5 x ) y = \text{cos}(5x) y = cos ( 5 x ) ? Answer: − 5 sin ( 5 x ) -5\text{sin}(5x) − 5 sin ( 5 x ) . Chain rule: − sin ( u ) × u ′ -\sin(u) \times u' − sin ( u ) × u ′ where u = 5 x u = 5x u = 5 x .
Flashcard 10: Differentiate y = e 3 x + 7 y = \text{e}^{3x+7} y = e 3 x + 7 . Answer: 3 e 3 x + 7 3\text{e}^{3x+7} 3 e 3 x + 7 . Chain rule: e u × u ′ e^u \times u' e u × u ′ where u = 3 x + 7 u = 3x + 7 u = 3 x + 7 .
Flashcard 11: Differentiate y = e ln ( x ) y = \text{e}^{\text{ln}(x)} y = e ln ( x ) . Answer: 1 x \frac{1}{x} x 1 . Simplifies to x x x since e ln ( x ) = x e^{\ln(x)} = x e l n ( x ) = x .
Flashcard 12: Differentiate y = cos − 1 ( 5 x ) y = \text{cos}^{-1}(5x) y = cos − 1 ( 5 x ) . Answer: − 5 sqrt ( 1 − 25 x 2 ) -\frac{5}{\text{sqrt}(1 - 25x^2)} − sqrt ( 1 − 25 x 2 ) 5 . Chain rule: − 1 1 − u 2 × u ′ -\frac{1}{\sqrt{1-u^2}} \times u' − 1 − u 2 1 × u ′ where u = 5 x u = 5x u = 5 x .
Flashcard 13: Find f ′ ( x ) f'(x) f ′ ( x ) if f ( x ) = cos ( e x ) f(x) = \text{cos}(\text{e}^x) f ( x ) = cos ( e x ) . Answer: − e x sin ( e x ) -\text{e}^x\text{sin}(\text{e}^x) − e x sin ( e x ) . Chain rule: − sin ( u ) × u ′ -\sin(u) \times u' − sin ( u ) × u ′ where u = e x u = e^x u = e x .
Flashcard 14: Find d / d x d/dx d / d x of y = cos ( e 2 x ) y = \text{cos}(\text{e}^{2x}) y = cos ( e 2 x ) . Answer: − 2 e 2 x sin ( e 2 x ) -2\text{e}^{2x}\text{sin}(\text{e}^{2x}) − 2 e 2 x sin ( e 2 x ) . Chain rule: − sin ( u ) × u ′ -\sin(u) \times u' − sin ( u ) × u ′ where u = e 2 x u = e^{2x} u = e 2 x .
Flashcard 15: Differentiate y = tan 2 ( 3 x ) y = \text{tan}^2(3x) y = tan 2 ( 3 x ) using the Chain Rule. Answer: 6 tan ( 3 x ) sec 2 ( 3 x ) 6\text{tan}(3x)\text{sec}^2(3x) 6 tan ( 3 x ) sec 2 ( 3 x ) . Chain rule: 2 u × u ′ 2u \times u' 2 u × u ′ where u = tan ( 3 x ) u = \tan(3x) u = tan ( 3 x ) .
Flashcard 16: Differentiate f ( x ) = e x 3 f(x) = \text{e}^{x^3} f ( x ) = e x 3 . Answer: 3 x 2 e x 3 3x^2\text{e}^{x^3} 3 x 2 e x 3 . Chain rule: e u × u ′ e^u \times u' e u × u ′ where u = x 3 u = x^3 u = x 3 .
Flashcard 17: Differentiate f ( x ) = cos 3 ( x ) f(x) = \text{cos}^3(x) f ( x ) = cos 3 ( x ) . Answer: − 3 cos 2 ( x ) sin ( x ) -3\text{cos}^2(x)\text{sin}(x) − 3 cos 2 ( x ) sin ( x ) . Chain rule: 3 cos 2 ( x ) × ( − sin ( x ) ) 3\cos^2(x) \times (-\sin(x)) 3 cos 2 ( x ) × ( − sin ( x )) .
Flashcard 18: Which function is the outer function in f ( g ( x ) ) = e 3 x + 7 f(g(x)) = \text{e}^{3x+7} f ( g ( x )) = e 3 x + 7 ? Answer: f ( u ) = e u f(u) = \text{e}^u f ( u ) = e u . The exponential function wraps the linear expression.
Flashcard 19: Evaluate d d x [ ln ( x 2 + 1 ) ] \frac{d}{dx}[\text{ln}(x^2 + 1)] d x d [ ln ( x 2 + 1 )] . Answer: 2 x x 2 + 1 \frac{2x}{x^2 + 1} x 2 + 1 2 x . Chain rule: 1 u × u ′ \frac{1}{u} \times u' u 1 × u ′ where u = x 2 + 1 u = x^2 + 1 u = x 2 + 1 .
Flashcard 20: Identify the outer function in f ( g ( x ) ) = tan ( x 2 3 ) f(g(x)) = \tan(\frac{x^2}{3}) f ( g ( x )) = tan ( 3 x 2 ) . Answer: f ( u ) = tan ( u ) f(u) = \tan(u) f ( u ) = tan ( u ) . The tangent function wraps the inner expression.
Flashcard 21: Differentiate y = ( 2 x + 5 ) 6 y = (2x + 5)^6 y = ( 2 x + 5 ) 6 . Answer: 12 ( 2 x + 5 ) 5 12(2x+5)^5 12 ( 2 x + 5 ) 5 . Power rule with chain rule: 6 u 5 × u ′ 6u^5 \times u' 6 u 5 × u ′ .
Flashcard 22: Differentiate g ( x ) = ln ( 4 x 2 + 1 ) g(x) = \text{ln}(4x^2 + 1) g ( x ) = ln ( 4 x 2 + 1 ) . Answer: 8 x 4 x 2 + 1 \frac{8x}{4x^2 + 1} 4 x 2 + 1 8 x . Chain rule: 1 u × u ′ \frac{1}{u} \times u' u 1 × u ′ where u = 4 x 2 + 1 u = 4x^2 + 1 u = 4 x 2 + 1 .
Flashcard 23: Differentiate e x 2 e^{x^2} e x 2 using the Chain Rule. Answer: 2 x e x 2 2xe^{x^2} 2 x e x 2 . Chain rule: derivative of e u e^u e u is e u e^u e u times u ′ u' u ′ .
Flashcard 24: Identify the inner function in f ( g ( x ) ) = tan ( x 2 3 ) f(g(x)) = \tan(\frac{x^2}{3}) f ( g ( x )) = tan ( 3 x 2 ) . Answer: g ( x ) = x 2 3 g(x) = \frac{x^2}{3} g ( x ) = 3 x 2 . The expression inside the tangent function.
Flashcard 25: State the formula for the Chain Rule. Answer: d y d x = d y d u × d u d x \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} d x d y = d u d y × d x d u . Fundamental chain rule formula for composite functions.
Flashcard 26: Find y ′ y' y ′ if y = tan ( ln ( x ) ) y = \text{tan}(\text{ln}(x)) y = tan ( ln ( x )) . Answer: 1 x cos 2 ( ln ( x ) ) \frac{1}{x\text{cos}^2(\text{ln}(x))} x cos 2 ( ln ( x )) 1 . Chain rule twice: sec 2 ( ln ( x ) ) × 1 x \sec^2(\ln(x)) \times \frac{1}{x} sec 2 ( ln ( x )) × x 1 .
Flashcard 27: Differentiate y = ( ln ( x ) ) 4 y = (\text{ln}(x))^4 y = ( ln ( x ) ) 4 using the Chain Rule. Answer: 4 ( ln ( x ) ) 3 x \frac{4(\text{ln}(x))^3}{x} x 4 ( ln ( x ) ) 3 . Power rule with chain rule: 4 u 3 × u ′ 4u^3 \times u' 4 u 3 × u ′ .
Flashcard 28: Find f ′ ( x ) f'(x) f ′ ( x ) if f ( x ) = sin ( x 3 ) f(x) = \text{sin}(x^3) f ( x ) = sin ( x 3 ) . Answer: 3 x 2 cos ( x 3 ) 3x^2 \text{cos}(x^3) 3 x 2 cos ( x 3 ) . Chain rule: cos ( u ) × u ′ \cos(u) \times u' cos ( u ) × u ′ where u = x 3 u = x^3 u = x 3 .
Flashcard 29: Differentiate f ( x ) = e sin ( x ) f(x) = \text{e}^{\text{sin}(x)} f ( x ) = e sin ( x ) . Answer: cos ( x ) e sin ( x ) \text{cos}(x)\text{e}^{\text{sin}(x)} cos ( x ) e sin ( x ) . Chain rule: e u × u ′ e^u \times u' e u × u ′ where u = sin ( x ) u = \sin(x) u = sin ( x ) .
Flashcard 30: What is the derivative of f ( g ( x ) ) f(g(x)) f ( g ( x )) using the Chain Rule? Answer: f ′ ( g ( x ) ) × g ′ ( x ) f'(g(x)) \times g'(x) f ′ ( g ( x )) × g ′ ( x ) . Chain rule applied to general composite function f ( g ( x ) ) f(g(x)) f ( g ( x )) .