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Featured 1 month ago

Write the expression as:

Now take the logarithm of this expression:

We can see that:

so we can ignore this term.

Focusing on the other addendum we have:

Substituting

(you can find the explanation here )

Putting this together we can see that:

Now note that:

so that:

and as

Featured 1 month ago

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

From the graph we can see that for the petal in Q1 then

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 #

Featured 1 month ago

The integral equals

By partial fractions:

We can write a system of equations.

Solving, we get

We can separate.

Let

We're going to use partial fractions again.

Write a system of equations

Solve to get

The expression becomes:

This can be readily integrated.

The logarithms can be combined.

Hopefully this helps!

Featured 1 month ago

The derivative of

Starting with:

use the chain rule to get:

use the chain rule once again on the remaining derivative:

Simplify:

Note that

Finally:

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

Rory.

Featured 3 weeks ago

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

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

#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

#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

Featured 3 weeks ago

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

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)#

#color(red)(darr)#

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

#color(red)(darr)#

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

2.

#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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