2. find the equation of the tangent line to the curve determined by the equation\n\n$$ y ^ { 3 } x + ( x - y…

2. find the equation of the tangent line to the curve determined by the equation\n\n$$ y ^ { 3 } x + ( x - y ) ^ { 4 } = 10 x - 5 y $$\n\nat the point (1,-1).\n\n$$\begin{array} { r } { y ^ { 3 } + x 3 y ^ { 2 } y ^ { prime } + 4 ( x - y ) ^ { 3 } ( 1 - y ^ { prime } ) = 10 - 5 y ^ { prime } } \\ { - 1 + 3 y ^ { prime } + 32 ( 1 - y ^ { prime } ) = 10 - 5 y ^ { prime } } \\ { 31 - 29 y ^ { prime } = 10 - 5 y ^ { prime } } \\ { 21 = 24 y ^ { prime } } \\ { \frac { 21 } { 24 } = y ^ { prime } = \frac { 7 } { 8 } } \\ { y ^ { prime } = \frac { 7 } { 8 } ( x - 1 ) - 1 } end{array}$$\n\n$$ y ^ { prime } = \frac { 7 } { 8 } ( x - 1 ) - 1 $$\n\n3. compute the derivative of $$ y = ( x ) ^ { cos x } $$ using logarithmic differentiation. take the natural log first of both sides of $$ y = ( x ) ^ { cos x } $$, use properties of natural log then differentiate and solve for $$ \frac { d y } { d x } $$.\n\n$$\begin{array} { l } { ln y = cos x ln x } \\ { \frac { 1 } { y } y ^ { prime } = cos ( \frac { 1 } { x } ) + ln x ( - sin x ) } \\ { y ^ { prime } = y ( \frac { cos x } { x } + - sin ( x ) ln x ) } end{array}$$\n\n$$ \frac { d y } { d x } = $$\n\n4. a crane with a 130 foot boom is lifting a box by increasing the angle it makes with the ground, $$ \theta $$. see picture below. if the angle $$ \theta $$ is increasing at a rate of $$ \frac { 1 } { 10 } $$ radians per minute, how fast is the box rising when the tip of the crane is 120 feet off the ground (h = 120)?\n\nft/min

2. find the equation of the tangent line to the curve determined by the equation\n\n$$ y ^ { 3 } x + ( x - y ) ^ { 4 } = 10 x - 5 y $$\n\nat the point (1,-1).\n\n$$\begin{array} { r } { y ^ { 3 } + x 3 y ^ { 2 } y ^ { prime } + 4 ( x - y ) ^ { 3 } ( 1 - y ^ { prime } ) = 10 - 5 y ^ { prime } } \\ { - 1 + 3 y ^ { prime } + 32 ( 1 - y ^ { prime } ) = 10 - 5 y ^ { prime } } \\ { 31 - 29 y ^ { prime } = 10 - 5 y ^ { prime } } \\ { 21 = 24 y ^ { prime } } \\ { \frac { 21 } { 24 } = y ^ { prime } = \frac { 7 } { 8 } } \\ { y ^ { prime } = \frac { 7 } { 8 } ( x - 1 ) - 1 } end{array}$$\n\n$$ y ^ { prime } = \frac { 7 } { 8 } ( x - 1 ) - 1 $$\n\n3. compute the derivative of $$ y = ( x ) ^ { cos x } $$ using logarithmic differentiation. take the natural log first of both sides of $$ y = ( x ) ^ { cos x } $$, use properties of natural log then differentiate and solve for $$ \frac { d y } { d x } $$.\n\n$$\begin{array} { l } { ln y = cos x ln x } \\ { \frac { 1 } { y } y ^ { prime } = cos ( \frac { 1 } { x } ) + ln x ( - sin x ) } \\ { y ^ { prime } = y ( \frac { cos x } { x } + - sin ( x ) ln x ) } end{array}$$\n\n$$ \frac { d y } { d x } = $$\n\n4. a crane with a 130 foot boom is lifting a box by increasing the angle it makes with the ground, $$ \theta $$. see picture below. if the angle $$ \theta $$ is increasing at a rate of $$ \frac { 1 } { 10 } $$ radians per minute, how fast is the box rising when the tip of the crane is 120 feet off the ground (h = 120)?\n\nft/min

Answer

Problem 2:

Explanation:

Step1: Differentiate implicitly

Differentiate (y^{3}x+(x - y)^{4}=10x-5y) with respect to (x). Using the product rule ((uv)^\prime = u^\prime v+uv^\prime) for (y^{3}x) ((u = y^{3}), (v=x)), we have (3y^{2}y^\prime x + y^{3}). Using the chain - rule ((u^{n})^\prime=nu^{n - 1}u^\prime) for ((x - y)^{4}) ((u=x - y), (n = 4)), we get (4(x - y)^{3}(1 - y^\prime)). The derivative of (10x-5y) is (10-5y^\prime). So, (3xy^{2}y^\prime+y^{3}+4(x - y)^{3}(1 - y^\prime)=10-5y^\prime).

Step2: Substitute (x = 1) and (y=-1)

Substitute (x = 1) and (y=-1) into (3xy^{2}y^\prime+y^{3}+4(x - y)^{3}(1 - y^\prime)=10-5y^\prime). First, calculate the values: (3\times1\times(-1)^{2}y^\prime+(-1)^{3}+4(1-(-1))^{3}(1 - y^\prime)=10-5y^\prime). (3y^\prime-1 + 4\times8(1 - y^\prime)=10-5y^\prime). (3y^\prime-1+32 - 32y^\prime=10-5y^\prime). (-29y^\prime + 31=10-5y^\prime). (-29y^\prime+5y^\prime=10 - 31). (-24y^\prime=-21), then (y^\prime=\frac{7}{8}).

Step3: Use the point - slope form

The point - slope form of a line is (y - y_{1}=m(x - x_{1})), where ((x_{1},y_{1})=(1,-1)) and (m = y^\prime=\frac{7}{8}). (y+1=\frac{7}{8}(x - 1)), or (y=\frac{7}{8}x-\frac{7}{8}-1=\frac{7}{8}x-\frac{15}{8})

Problem 3:

Explanation:

Step1: Take the natural logarithm

Given (y = x^{\cos x}), take (\ln) of both sides: (\ln y=\cos x\ln x).

Step2: Differentiate both sides

Differentiate (\ln y=\cos x\ln x) with respect to (x). Using the product rule ((uv)^\prime = u^\prime v+uv^\prime) ((u=\cos x), (v = \ln x)), we have (\frac{1}{y}y^\prime=-\sin x\ln x+\frac{\cos x}{x}).

Step3: Solve for (y^\prime)

Multiply both sides by (y=x^{\cos x}) to get (y^\prime=x^{\cos x}\left(\frac{\cos x}{x}-\sin x\ln x\right))

Problem 4:

Explanation:

Step1: Establish the relationship

We know that (h = 130\sin\theta) (from the right - triangle, (\sin\theta=\frac{h}{130})).

Step2: Differentiate with respect to time (t)

Differentiate (h = 130\sin\theta) with respect to (t) using the chain - rule (\frac{dh}{dt}=130\cos\theta\frac{d\theta}{dt}). When (h = 120), (\sin\theta=\frac{120}{130}=\frac{12}{13}), then (\cos\theta=\sqrt{1-\sin^{2}\theta}=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\frac{5}{13}). Given (\frac{d\theta}{dt}=\frac{1}{10}) radians per minute.

Step3: Calculate (\frac{dh}{dt})

Substitute (\cos\theta=\frac{5}{13}) and (\frac{d\theta}{dt}=\frac{1}{10}) into (\frac{dh}{dt}=130\cos\theta\frac{d\theta}{dt}). (\frac{dh}{dt}=130\times\frac{5}{13}\times\frac{1}{10}=5)

Answer:

  • Problem 2: (y=\frac{7}{8}x-\frac{15}{8})
  • Problem 3: (y^\prime=x^{\cos x}\left(\frac{\cos x}{x}-\sin x\ln x\right))
  • Problem 4: (5) ft/min