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Answer:

#lim_(x->oo) (x/(2+x))^(x-2) = 1/e^2#

Explanation:

Write the expression as:

#(x/(2+x))^(x-2) = 1/((x/(2+x))^(-(x-2))) = ((x+2)/x)^(2-x) = (1+2/x)^(2-x)#

Now take the logarithm of this expression:

#ln((x/(2+x))^(x-2)) = (2-x)ln(1+2/x) = 2ln(1+2/x)-xln(1+2/x)#

We can see that:

#lim_(x->oo) 2ln(1+2/x) = 2ln(1) = 0#

so we can ignore this term.

Focusing on the other addendum we have:

#xln(1+2/x) = 2 ln(1+2/x)/(2/x)#

Substituting #y=2/x# we have #lim_(x->oo) y(x) = 0# so that:

#lim_(x->oo) ln(1+2/x)/(2/x) = lim_(y->0) ln (1+y)/y = 1#

(you can find the explanation here )

Putting this together we can see that:

#lim_(x->oo) ln((x/(2+x))^(x-2)) = lim_(x->oo) 2ln(1+2/x)-xln(1+2/x) = 0-2*1 =-2#

Now note that:

#e^(ln((x/(2+x))^(x-2)))=(x/(2+x))^(x-2)#

so that:

#lim_(x->oo) (x/(2+x))^(x-2) = lim_(x->oo) (e^(ln((x/(2+x))^(x-2))))#

and as #e^x# is continuous in all of #RR#:

#lim_(x->oo) (x/(2+x))^(x-2) = e^(lim_(x->oo) (ln((x/(2+x))^(x-2)))) = e^(-2) = 1/e^2#

Answer:

#(9pi)/2 #

Explanation:

Area in polar coordinates is given by:

# A = int_alpha^beta 1/2 r^2 \ d theta #

The first step is to plot the polar curve to establish the appropriate range of #theta#

enter image source here

From the graph we can see that for the petal in Q1 then #theta in [0, pi/2]#

Hence,

# A = int_0^(pi/2) 1/2 (6sin2theta)^2 \ d theta #
# \ \ \ = 18 \ int_0^(pi/2) sin^2 2theta \ d theta #
# \ \ \ = 18 \ int_0^(pi/2) 1/2(1-cos 4 theta) \ d theta #
# \ \ \ = 9 \ int_0^(pi/2) 1-cos 4 theta \ d theta #
# \ \ \ = 9 \ [ theta -1/4sin 4 theta ]_0^(pi/2) #
# \ \ \ = 9 \ {(pi/2-1/4sin(2pi)) - (0)} #
# \ \ \ = (9pi)/2 #

Answer:

The integral equals #x - 2ln|e^x/(e^x + 1)| + C#

Explanation:

By partial fractions:

#A/1 + B/(e^x + 1) = e^x - 2 + 1#

#A(e^x + 1) + B = e^x- 1#

#Ae^x + A + B = e^x- 1#

#Ae^x + (A + B) = e^x - 1#

We can write a system of equations.

#{(A = 1), (A + B = -1):}#

Solving, we get #A = 1# and #B = -2#. The integral becomes:

#=>int 1 - 2/(e^x + 1)dx#

We can separate.

#=>int 1dx - 2int1/(e^x + 1)dx#

Let #u = e^x#. Then #du = e^xdx#. Then #dx = (du)/e^x = (du)/u#

#=>int 1dx - 2int 1/((u + 1)u)du#

We're going to use partial fractions again.

#A/(u + 1) + B/u = 1#

#Au + Bu + B = 1#

Write a system of equations

#{(A + B = 0), (B = 1):}#

Solve to get #A = -1# and #B = 1#.

The expression becomes:

#=>int1dx - 2(int 1/udu - int1/( u + 1)du)#

#=>int1dx - 2int1/udu + 2int 1/(u + 1)du#

This can be readily integrated.

#=>x - 2ln|u| + 2ln|u +1| + C#

#=>x - 2(ln|u| - ln|u + 1|) + C#

The logarithms can be combined.

#=>x - 2ln|u/(u + 1)| + C#

#=>x - 2ln|e^x/(e^x + 1)| + C#

Hopefully this helps!

Answer:

The derivative of #-ln(x - (x^2+1)^(1/2))# with respect to x is #1/sqrt(x^2 +1).# To solve this use the chain rule carefully. See the explanation for details.

Explanation:

Starting with:

#d / dx (-ln(x - (x^2+1)^(1/2)))#

use the chain rule to get:

#d / dx (-ln(x - (x^2+1)^(1/2)))#

#= -1/(x-(x^2+1)^(1/2)) *d/dx (x - (x^2+1)^(1/2))#

use the chain rule once again on the remaining derivative:

#= -1/(x-(x^2+1)^(1/2)) * (1 - 1/2(x^2+1)^(-1/2)(2x))#

Simplify:

#= (x/(x^2+1)^(1/2) -1 )/(x-(x^2+1)^(1/2))#

Note that #1=(x^2+1)^(1/2) / (x^2+1)^(1/2)#, then substitute this for 1:

#= (x/(x^2+1)^(1/2) - (x^2+1)^(1/2) / (x^2+1)^(1/2) )/(x-(x^2+1)^(1/2))#

#= ((x - (x^2+1)^(1/2)) / (x^2+1)^(1/2) )/(x-(x^2+1)^(1/2))#

#= ((x - (x^2+1)^(1/2)) / (x^2+1)^(1/2) ) divide (x-(x^2+1)^(1/2))#

#= ((x - (x^2+1)^(1/2)) / (x^2+1)^(1/2) ) * 1/(x-(x^2+1)^(1/2))#

#= (x - (x^2+1)^(1/2)) / (x-(x^2+1)^(1/2) ) * 1/((x^2+1)^(1/2))#

#= 1 * 1/((x^2+1)^(1/2)) #

#= 1/((x^2+1)^(1/2)) = 1/sqrt((x^2+1)).#

Finally:

#d / dx (-ln(x - (x^2+1)^(1/2))) = 1/sqrt((x^2+1)).#

If you have any questions about the use of the chain rule or any other part of this solution, then please ask.

Rory.

Answer:

#P'(16) = 75/32#

Explanation:

The idea here is that you can find the rate of change of the pollution with respect to time by taking the first derivative of your function #P(t)#.

So this is pretty much an exercise in finding the derivative of the function

#P(t) = (t^(1/4) + 3)^3#

which can be calculated by using the chain rule and the power rule. Keep in mind that you have

#color(blue)(ul(color(black)(d/(dt)[u(t)^n] = n * u(t)^(n-1) * d/(dt)[u(t)])))#

In your case,

#{(u(t) = t^(1/4) + 3), (n = 3) :}#

This means that the derivate of #P(t)# will be

#overbrace(d/d(dt)[P(t)])^(color(blue)(=P^(')(t))) = 3 * (t^(1/4) + 3)^2 * d/(dt)(t^(1/4) + 3)#

#P^'(t) = 3 * (t^(1/4) + 3)^2 * 1/4 * t^((1/4 - 1))#

#P^'(t) = 3/4 * t^(-3/4) * (t^(1/4) + 3)^2#

Now all you have to do is plug in #16# for #t# to find

#P^'(t) = 3/4 * 16^(-3/4) * (16^(1/4) + 3)^2#

Since you know that

#16 = 2^4#

you can rewrite the equation as

#P^'(16) = 3/4 * 2^[[4 * (-3/4)]] * [2^((4 * 1/4)) + 3]^2#

#P^'(16) = 3/4 * 2^(-3) * (2 + 3)^2#

#P^'(16) = 3/4 * 1/8 * 25#

#P^'(16) = 75/32#

And there you have it -- the rate at which the pollution changes after #16# years.

Answer:

#y=3/(2(4+3x)^(1/2))#

Explanation:

Recall the formula for chain rule:

#color(blue)(bar(ul(|color(white)(a/a)dy/dx=dy/(du)(du)/dxcolor(white)(a/a)|)))# or #color(blue)(bar(ul(|color(white)(a/a)f'(x)=g'[h(x)]h'(x)color(white)(a/a)|)))#

and the formula for power rule:

#color(blue)(bar(ul(|color(white)(a/a)d/dx(x^n)=nx^(n-1)color(white)(a/a)|)))#

To start, recognize the inside and outside functions of #y=color(green)(sqrt(color(darkorange)(4+3x)))#.

Inside function: #y=color(darkorange)(4+3x)#
Outside function: #y=color(green)(sqrt(a))#

How to Differentiate Using Chain Rule

#1#. Take the derivative of the outside function, #y=sqrt(a)#, but replace the #a# with the inside function, #4+3x#.

#2#. Multiply by the derivative of the inside function, #4+3x#.

Applying Chain Rule
1. The derivative of the outside function, #y=sqrt(a)#, would be #1/2a^(-1/2)#, using the power rule. However, #a# needs to be replaced by the inside function, so it becomes #1/2(4+3x)^(-1/2)#.

#y=sqrt(a)#

#color(red)(darr)#

#y=1/2a^(-1/2)#

#color(red)(darr)#

#y=1/2(4+3x)^(-1/2)#

2.#color(white)(i)#Then we need to multiply by the derivative of the inside function, #4+3x#, which becomes #3# using the power rule.

#y=1/2(4+3x)^(-1/2)#

#color(red)(darr)#

#y=1/2(4+3x)^(-1/2)(3)#

#color(green)(bar(ul(|color(white)(a/a)y=3/(2(4+3x)^(1/2))color(white)(a/a)|)))#

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