University:
California State University, Northridge
Course:
MATH 351 | Differential Equations
Academic year:

2024
Views:

312

Pages:

2
Author:

Javon Neal

Question 1 Finding the derivative of sin ⁡ 2 𝑥 3 3 sin2x using only the definition of derivative. Answer Let 𝑎 = sin ⁡ 2 𝑥 + ℎ 3 a= 3 sin2x+h and 𝑏 = sin ⁡ 2 𝑥 3 b= 3 sin2x . Using the identity 𝑎 3 − 𝑏 3 = ( 𝑎 − 𝑏 ) ( 𝑎 2 + 𝑎 𝑏 + 𝑏 2 ) a 3 −b 3 =(a−b)(a 2 +ab+b 2 ), we have: sin ⁡ ( 2 𝑥 + ℎ ) 3 − sin ⁡ 2 𝑥 3 ℎ = sin ⁡ ( 2 𝑥 + ℎ ) − sin ⁡ ( 2 𝑥 ) ℎ ⋅ 1 𝑎 2 + 𝑎 𝑏 + 𝑏 2 . h 3 sin(2x+h) − 3 sin2x = h sin(2x+h)−sin(2x) ⋅ a 2 +ab+b 2 1 . Using the identity for the sine difference: sin ⁡ ( 2 𝑥 + ℎ ) − sin ⁡ ( 2 𝑥 ) = 2 cos ⁡ ( 2 𝑥 + ℎ / 2 ) sin ⁡ ( ℎ / 2 ) . sin(2x+h)−sin(2x)=2cos(2x+h/2)sin(h/2). Taking the limit as ℎ → 0 h→0: lim ⁡ ℎ → 0 sin ⁡ ( 2 𝑥 + ℎ ) − sin ⁡ ( 2 𝑥 ) ℎ = 2 cos ⁡ ( 2 𝑥 ) . h→0 lim h sin(2x+h)−sin(2x) =2cos(2x). Thus, the derivative of sin ⁡ 2 𝑥 3 3 sin2x is: 𝑑 𝑑 𝑥 ( sin ⁡ 2 𝑥 3 ) = 2 cos ⁡ ( 2 𝑥 ) 3 sin ⁡ 2 / 3 ( 2 𝑥 ) . dx d ( 3 sin2x )= 3sin 2/3 (2x) 2cos(2x) . Question 2 How do you find the equation of a tangent line to a curve? Answer To find the tangent line to a curve: Let 𝑓 ( 𝑥 ) f(x) be the function for the curve. Find 𝑓 ′ ( 𝑥 ) f ′ (x), the derivative of 𝑓 ( 𝑥 ) f(x). At 𝑥 = 𝑎 x=a, the equation of the tangent line is: 𝑇 ( 𝑥 ) = 𝑓 ( 𝑎 ) + 𝑓 ′ ( 𝑎 ) ( 𝑥 − 𝑎 ) . T(x)=f(a)+f ′ (a)(x−a). This is also the formula for linear approximation functions. Question 3 How do you find the slope of a tangent line using secant lines? Answer The slope of a tangent line can be approximated using the slope of a secant line: 𝑚 = 𝑓 ( 𝑎 + ℎ ) − 𝑓 ( 𝑎 ) ℎ . m= h f(a+h)−f(a) . As ℎ h approaches 0, the slope of the secant line becomes the slope of the tangent line: 𝑚 = lim ⁡ ℎ → 0 𝑓 ( 𝑎 + ℎ ) − 𝑓 ( 𝑎 ) ℎ . m= h→0 lim h f(a+h)−f(a) . Question 4 How do you find a tangent line parallel to a secant line? Answer The Mean Value Theorem is used to find a tangent line parallel to a secant line. The theorem states that if a function 𝑓 ( 𝑥 ) f(x) is continuous and differentiable on [ 𝑎 , 𝑏 ] [a,b], then: 𝑓 ′ ( 𝑐 ) = 𝑓 ( 𝑏 ) − 𝑓 ( 𝑎 ) 𝑏 − 𝑎 , for some 𝑐 ∈ ( 𝑎 , 𝑏 ) . f ′ (c)= b−a f(b)−f(a) ,for some c∈(a,b). Example: For 𝑓 ( 𝑥 ) = − 𝑥 3 f(x)=−x 3 , if the secant line connects ( − 2 , − 8 ) (−2,−8) and ( 2 , − 8 ) (2,−8): The slope of the secant line is: 𝑚 = − 8 − ( − 8 ) 2 − ( − 2 ) = − 4. m= 2−(−2) −8−(−8) =−4. To find the tangent line parallel to this secant, solve 𝑓 ′ ( 𝑥 ) = − 4 f ′ (x)=−4: 𝑓 ′ ( 𝑥 ) = − 3 𝑥 2 = − 4. f ′ (x)=−3x 2 =−4. This gives 𝑥 = ± 4 3 x=± 3 4 . Question 5 How does the derivative relate to the tangent line? Answer The derivative of a function at a given point provides the slope of the tangent line to the curve at that point. For 𝑦 = 𝑓 ( 𝑥 ) y=f(x), the slope of the tangent line at 𝑥 0 x 0 is: slope = 𝑓 ′ ( 𝑥 0 ) . slope=f ′ (x 0 ).