2023: section three\n1. differentiate the following functions. you do not need to simplify your answer.\ni…

2023: section three\n1. differentiate the following functions. you do not need to simplify your answer.\ni. (f(x)=\tan2x) sl = 1\nii. (y = lnsqrt{2x + 1}) sl = 2\niii. (g(x)=\frac{e^{2x}}{(x^{2}+1)^{10}}) sl = 3\n2. find (lim_{x\rightarrowinfty}\frac{1 + 2x^{2}-x^{3}}{3x^{3}-2x + 1}) sl = 2\n3. the graph of a piece - wise function (f(x)) is shown below.\ni. find (lim_{x\rightarrow2}f(x)) sl = 1\nii. sketch the graph of the derivative of (f(x)). sl = 2\n4. use implicit differentiation to find (\frac{dy}{dx}) if (x^{2}+y^{2}-2y = 1) sl = 2\n5. the curve traced by a point on a circle as it rolls on a straight line has parametric equations (x=\theta-sin\theta,y = 1-cos\theta). find (\frac{d^{2}y}{dx^{2}}) in terms of (\theta). sl = 3\n6. a 10 m ladder is leaning against a vertical wall. the angle, (\theta), formed by the ladder and the ground changes if the top of the ladder slides down the wall at a constant rate of 2 m/s, what rate of the angle (\theta) changing when the top of the ladder is 6 m above the ground sl = 3\n7. a cone is inscribed in a sphere of radius 3 cm, centered at o. the height of the cone is (h) and the radius of the base is (r). find the height of the cone so that the cone has a maximum volume. sl = 4

2023: section three\n1. differentiate the following functions. you do not need to simplify your answer.\ni. (f(x)=\tan2x) sl = 1\nii. (y = lnsqrt{2x + 1}) sl = 2\niii. (g(x)=\frac{e^{2x}}{(x^{2}+1)^{10}}) sl = 3\n2. find (lim_{x\rightarrowinfty}\frac{1 + 2x^{2}-x^{3}}{3x^{3}-2x + 1}) sl = 2\n3. the graph of a piece - wise function (f(x)) is shown below.\ni. find (lim_{x\rightarrow2}f(x)) sl = 1\nii. sketch the graph of the derivative of (f(x)). sl = 2\n4. use implicit differentiation to find (\frac{dy}{dx}) if (x^{2}+y^{2}-2y = 1) sl = 2\n5. the curve traced by a point on a circle as it rolls on a straight line has parametric equations (x=\theta-sin\theta,y = 1-cos\theta). find (\frac{d^{2}y}{dx^{2}}) in terms of (\theta). sl = 3\n6. a 10 m ladder is leaning against a vertical wall. the angle, (\theta), formed by the ladder and the ground changes if the top of the ladder slides down the wall at a constant rate of 2 m/s, what rate of the angle (\theta) changing when the top of the ladder is 6 m above the ground sl = 3\n7. a cone is inscribed in a sphere of radius 3 cm, centered at o. the height of the cone is (h) and the radius of the base is (r). find the height of the cone so that the cone has a maximum volume. sl = 4

Answer

Explanation:

1. Differentiate (f(x)=\tan(2x))

Use the chain - rule. If (y = \tan(u)) and (u = 2x), the derivative of (\tan(u)) with respect to (u) is (\sec^{2}(u)), and the derivative of (u = 2x) with respect to (x) is (2). [f^\prime(x)=2\sec^{2}(2x)]

2. Differentiate (y = \ln\sqrt{2x + 1})

First, rewrite (y=\ln(2x + 1)^{\frac{1}{2}}=\frac{1}{2}\ln(2x + 1)). Then, using the chain - rule, if (y=\frac{1}{2}\ln(u)) and (u = 2x+1), the derivative of (\ln(u)) with respect to (u) is (\frac{1}{u}), and the derivative of (u = 2x + 1) with respect to (x) is (2). [y^\prime=\frac{1}{2}\times\frac{2}{2x + 1}=\frac{1}{2x+1}]

3. Differentiate (g(x)=\frac{e^{2x}}{(x^{2}+1)^{10}})

Use the quotient - rule ((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}), where (u = e^{2x}), (u^\prime=2e^{2x}), (v=(x^{2}+1)^{10}), and (v^\prime = 10(x^{2}+1)^{9}\times2x). [g^\prime(x)=\frac{2e^{2x}(x^{2}+1)^{10}-e^{2x}\times20x(x^{2}+1)^{9}}{(x^{2}+1)^{20}}]

4. Find (\lim_{x\rightarrow\infty}\frac{1 + 2x^{2}-x^{3}}{3x^{3}-2x + 1})

Divide both the numerator and denominator by (x^{3}): [ \begin{align*} \lim_{x\rightarrow\infty}\frac{1 + 2x^{2}-x^{3}}{3x^{3}-2x + 1}&=\lim_{x\rightarrow\infty}\frac{\frac{1}{x^{3}}+\frac{2}{x}-1}{3-\frac{2}{x^{2}}+\frac{1}{x^{3}}}\ &=\frac{0 + 0-1}{3-0 + 0}=-\frac{1}{3} \end{align*} ]

5. For the piece - wise function (f(x)) (graph given):

i. Find (\lim_{x\rightarrow2}f(x))

Examine the left - hand limit and right - hand limit as (x\rightarrow2) from the graph. If the left - hand limit (\lim_{x\rightarrow2^{-}}f(x)) and the right - hand limit (\lim_{x\rightarrow2^{+}}f(x)) are equal, then (\lim_{x\rightarrow2}f(x)) exists.

ii. Sketch the graph of (f^\prime(x))

Analyze the slope of (f(x)) at different intervals from the graph of (f(x)). Where (f(x)) is increasing, (f^\prime(x)>0); where (f(x)) is decreasing, (f^\prime(x)<0); and at the points of horizontal tangency, (f^\prime(x) = 0).

6. Use implicit differentiation for (x^{2}+y^{2}-2y = 1)

Differentiate both sides with respect to (x): [2x + 2y\frac{dy}{dx}-2\frac{dy}{dx}=0] [2y\frac{dy}{dx}-2\frac{dy}{dx}=-2x] [\frac{dy}{dx}=\frac{-2x}{2y - 2}=\frac{x}{1 - y}]

7. For parametric equations (x=\theta-\sin\theta), (y = 1-\cos\theta)

First, find (\frac{dy}{d\theta}=\sin\theta) and (\frac{dx}{d\theta}=1-\cos\theta). Then (\frac{dy}{dx}=\frac{\sin\theta}{1 - \cos\theta}). To find (\frac{d^{2}y}{dx^{2}}), use the formula (\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{d\theta}(\frac{dy}{dx})}{\frac{dx}{d\theta}}). (\frac{d}{d\theta}(\frac{\sin\theta}{1 - \cos\theta})=\frac{\cos\theta(1 - \cos\theta)-\sin^{2}\theta}{(1 - \cos\theta)^{2}}=\frac{\cos\theta - 1}{(1 - \cos\theta)^{2}}=-\frac{1}{1 - \cos\theta}) So (\frac{d^{2}y}{dx^{2}}=-\frac{1}{(1 - \cos\theta)^{2}})

8. For the ladder problem

Let (y) be the height of the top of the ladder on the wall. We know (y = 6) m, the length of the ladder (L = 10) m. By the Pythagorean theorem, (x^{2}+y^{2}=100), and (\sin\theta=\frac{y}{10}). Differentiate (\sin\theta=\frac{y}{10}) with respect to time (t): (\cos\theta\frac{d\theta}{dt}=\frac{1}{10}\frac{dy}{dt}). When (y = 6), (x=\sqrt{100 - 36}=8), so (\cos\theta=\frac{8}{10}) and (\frac{dy}{dt}=-2) m/s. (\frac{8}{10}\frac{d\theta}{dt}=\frac{1}{10}\times(-2)) (\frac{d\theta}{dt}=-\frac{1}{4}) rad/s

9. For the cone inscribed in a sphere

The radius of the sphere (R = 3) cm. Using the Pythagorean relationship (r^{2}=R^{2}-(h - R)^{2}=9-(h - 3)^{2}=6h - h^{2}). The volume of the cone (V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi(6h - h^{2})h=\frac{1}{3}\pi(6h^{2}-h^{3})). Differentiate (V) with respect to (h): (V^\prime=\frac{1}{3}\pi(12h - 3h^{2})). Set (V^\prime = 0): (12h-3h^{2}=0), (3h(4 - h)=0). So (h = 4) cm.

Answer:

  1. i. (2\sec^{2}(2x)) ii. (\frac{1}{2x + 1}) iii. (\frac{2e^{2x}(x^{2}+1)^{10}-20xe^{2x}(x^{2}+1)^{9}}{(x^{2}+1)^{20}})
  2. (-\frac{1}{3})
  3. i. Value of limit from graph analysis ii. Sketch of derivative graph from original graph analysis
  4. (\frac{x}{1 - y})
  5. (-\frac{1}{(1 - \cos\theta)^{2}})
  6. (-\frac{1}{4}) rad/s
  7. (h = 4) cm