Given x=et and y=te-t, find dydx and dydx2 in terms of t.

1 - t e 2 t , d 2 y d x 2 = t - 2 e 3 t

Given x = e t and y = t e - t , find d y d x and d y d x 2 in terms of t .

1 - t e 2 t , d 2 y d x 2 = t - 2 e 3 t

t + 1 e 2 t , d 2 y d x 2 = 2 - t e 3 t

t - 1 e 2 t , d 2 y d x 2 = 2 + t e 3 t

1 + t e 2 t , d 2 y d x 2 = t + 2 e 3 t

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0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

2. Intro to Derivatives

Derivatives as Functions

Calculus

2. Intro to Derivatives

Derivatives as Functions

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Multiple Choice

Given $x = e^{t}$ and $y = t e^{- t}$ , find $\frac{d y}{d x}$ and ${\frac{d y}{d x}}^{2}$ in terms of $t$ .

\frac{1 - t}{e^{2 t}}

\frac{d^{2} y}{d x^{2}}

\frac{t - 2}{e^{3 t}}

\frac{t + 1}{e^{2 t}}

\frac{d^{2} y}{d x^{2}}

\frac{2 - t}{e^{3 t}}

\frac{t - 1}{e^{2 t}}

\frac{d^{2} y}{d x^{2}}

\frac{2 + t}{e^{3 t}}

\frac{1 + t}{e^{2 t}}

\frac{d^{2} y}{d x^{2}}

\frac{t + 2}{e^{3 t}}

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Verified step by step guidance

Step 1: Start by finding dx/dt and dy/dt. Since x = e^t, differentiate x with respect to t to get dx/dt = e^t. For y = t e^{-t}, use the product rule to differentiate y with respect to t: dy/dt = d(t)/dt * e^{-t} + t * d(e^{-t})/dt = 1 * e^{-t} + t * (-e^{-t}) = e^{-t} - t e^{-t}.

Step 2: To find dy/dx, use the chain rule: dy/dx = (dy/dt) / (dx/dt). Substitute dy/dt = e^{-t} - t e^{-t} and dx/dt = e^t into the formula: dy/dx = (e^{-t} - t e^{-t}) / e^t. Simplify the expression by combining the exponents: dy/dx = (1 - t) / e^{2t}.

Step 3: To find d^2y/dx^2, first compute d(dy/dx)/dt. Differentiate dy/dx = (1 - t) / e^{2t} with respect to t using the quotient rule: d(dy/dx)/dt = [(d(1 - t)/dt) * e^{2t} - (1 - t) * d(e^{2t}/dt)] / (e^{2t})^2. Simplify each term: d(1 - t)/dt = -1, and d(e^{2t})/dt = 2 e^{2t}.

Step 4: Substitute the simplified derivatives into the formula for d(dy/dx)/dt: d(dy/dx)/dt = [(-1) * e^{2t} - (1 - t) * 2 e^{2t}] / e^{4t}. Combine terms in the numerator: d(dy/dx)/dt = [-e^{2t} - 2 e^{2t} + 2t e^{2t}] / e^{4t} = (-3 + 2t) / e^{2t}.

Step 5: To find d^2y/dx^2, divide d(dy/dx)/dt by dx/dt: d^2y/dx^2 = [(-3 + 2t) / e^{2t}] / e^t. Simplify the expression by combining the exponents: d^2y/dx^2 = (2t - 3) / e^{3t}.