Answer to Question #202304 in Calculus for Zinzi Nkonco

Question

1. Use the method of implicit differentiation to determine the derivatives of the following:

a. xsin y + ysin x =1

b. tan(x-y) =y/1+x^2

C. √x+y = x^4 + y^4

d.y + xcos y = x^2 y

2. Find the number "c" that satisfy the Mean Value Theorem (M.V.T) on the given intervals:

a. f(x)= e^-x ; [ 0,2 ]

b. f(x)=x/x+2 ; [ 1, π ]

3.Determine the equation of the tangent and normal at the given points:

a. y + xcos y = x^2 y ; [ 1, π/2 ]

b. h(x) = 2/√x^2 + 1 ; at x=1

Answer 1

1.

a.

$$x\sin y + y\sin x =1$$

Diffirentiate both sides with respect to $x$

$$\sin y+(x\cos y)y'+(\sin x)y'+y\cos x=0$$

$$y'=-\dfrac{\sin y+y\cos x}{\sin x+x\cos y}$$

b.

$$\tan(x-y) =\dfrac{y}{1+x^2}$$

$$(1+x^2)\tan(x-y)=y$$

Diffirentiate both sides with respect to $x$

$$2x\tan(x-y)+\dfrac{(1+x^2)(1-y')}{\cos ^2(x-y)}=y'$$

$$y'=\dfrac{2x\tan(x-y)+1+x^2}{1+x^2+\cos ^2(x-y)}$$

c.

$$\sqrt{x+y} = x^4 + y^4$$

Diffirentiate both sides with respect to $x$

$$\dfrac{(1+y')}{2\sqrt{x+y}}=4x^3+4y^3y'$$

$$y'=\dfrac{8x^3\sqrt{x+y}-1}{1-8y^3\sqrt{x+y}}$$

d.

$$y + x\cos y = x^2 y$$

Diffirentiate both sides with respect to $x$

$$y'+\cos y-(x\sin y)y'=2xy+x^2 y'$$

$$y'=\dfrac{2xy-\cos y}{1-x\sin y-x^2}$$

2.

Find the number "c" that satisfy the Mean Value Theorem (M.V.T) on the given intervals:

a. $f(x)=e^{-x}, [0, 2]$

$$f'(x)=-e^{-x}$$

$$f'(c)=-e^{-c}, c\in(0, 2)$$

$$f(0)=e^{-0}=1$$

$$f(2)=e^{-2}$$

Use the Mean Value Theorem

$$f'(c)=\dfrac{f(2)-f(0)}{2-0}$$

$$-e^{-c}=\dfrac{e^{-2}-1}{2-0}$$

$$e^c=\dfrac{2}{1-e^{-2}}$$

$$c=\ln(\dfrac{2}{1-e^{-2}})$$

b. $f(x)=\dfrac{x}{x+2}, [1, \pi]$

$$f'(x)=\dfrac{x+2-x}{(x+2)^2}=\dfrac{2}{(x+2)^2}$$

$$f'(c)=\dfrac{2}{(c+2)^2}, c\in(1, \pi)$$

$$f(1)=\dfrac{1}{1+2}=\dfrac{1}{3}$$

$$f(\pi)=\dfrac{\pi}{\pi+2}$$

Use the Mean Value Theorem

$$f'(c)=\dfrac{f(\pi)-f(1)}{\pi-1}$$

$$\dfrac{2}{(c+2)^2}=\dfrac{\dfrac{\pi}{\pi+2}-\dfrac{1}{3}}{\pi-1}, c\in(1, \pi)$$

$$\dfrac{2}{(c+2)^2}=\dfrac{3\pi-\pi-2}{3(\pi-1)(\pi+2)}$$

$$\dfrac{1}{(c+2)^2}=\dfrac{1}{3(\pi+2)}$$

Since $c\in (1, \pi),$ then

$$c=\sqrt{3(\pi+2)}-2$$

3.Determine the equation of the tangent and normal at the given points:

a.

$$y'+\cos y-(x\sin y)y'=2xy+x^2y'$$

$$y'=\dfrac{\cos y-2xy}{x\sin y+x^2-1}$$

Point $(1, \pi/2)$

$$slope_1=m_1=\dfrac{\cos (\pi/2)-2(1)(\pi/2)}{1\sin (\pi/2)+(1)^2-1}=-\pi$$

The equation of the tangent line in point-slope form

$$y-\pi/2=-\pi(x-1)$$

The equation of the tangent line in slope-intercept form

$$y=-\pi x+3\pi/2$$

$$slope_2=m_2=-\dfrac{1}{m_1}=\dfrac{1}{\pi}$$

The equation of the normal line in point-slope form

$$y-\pi/2=\dfrac{1}{\pi}(x-1)$$

The equation of the normal line in slope-intercept form

$$y=\dfrac{1}{\pi} x+\pi/2-\dfrac{1}{\pi}$$

b.

$$h'(x)=-\dfrac{2x}{(x^2+1)^{3/2}}$$

$x=1$

$$slope_1=m_1=h'(1)=-\dfrac{2(1)}{(1^2+1)^{3/2}}=-\dfrac{\sqrt{2}}{2}$$

$$h(1)=\dfrac{2}{\sqrt{1^2+1}}=\sqrt{2}$$

The equation of the tangent line in point-slope form

$$y-\sqrt{2}=-\dfrac{\sqrt{2}}{2}(x-1)$$

The equation of the tangent line in slope-intercept form

$$y=-\dfrac{\sqrt{2}}{2} x+\dfrac{3\sqrt{2}}{2}$$

$$slope_2=m_2=-\dfrac{1}{m_1}=\sqrt{2}$$

The equation of the normal line in point-slope form

$$y-\sqrt{2}=\sqrt{2}(x-1)$$

The equation of the normal line in slope-intercept form

$$y=\sqrt{2}x$$