Implicit Differentiation
Let the function y(x) be defined implicitly on 0<x<0 by ln(x/e^y)=cos(y).
What is the derivative of this function?
By using the chain rule and applying the inverses of both the natural logarithm and of cosine, the resulting derivative should be:
y'=1/(x-sin(y)x)
Let f(x)=sin(x^4)
Is the function increasing or decreasing at x=10?
f'(x)=4x^3(cos(x^4))
=4(10)^3(cos(10)^4)
=-3808.62
The derivative at x=10 is negative, therefore the function should be decreasing.
Why do people appear bright until they speak?
Because light travels faster than sound
Let f be a differentiable function that's
-defined everywhere
-has x=13 as its only x intercept
-has x=1 as a local max which is its only critical point
Let g(x) = (1/f(x^2-3))^(2)+3
How many zeros does g(x) have?
g(x) should have no zeros; this is is because (1/f(x^2-3))^2 cannot equal -3 due to the square (this part of the equation must be equal to -3 in order for g(x) to equal 0).
The best quadratic approximation to y=2lnx is given in the form displayed below
y=ax^2+bx+c
where a, b, and c are constants. Knowing that this approximation has the same first and second derivatives as y=2lnx at x=1, solve for a.
First, solve for the second derivative of the approximation and of the original function. Then, set each respective second derivative to each other to solve for a; your answer should be -1.
Let the function y(x) be defined implicitly on 0<x<0 by ln(x/e^y)=cos(y).
Find the y coordinate where this curve y(x) has a vertical tangent line for 0<y<pi.
Find the derivative of y(x) first, then set this equation to undefined (this is because we are trying to find the x value where y(x) has a vertical tangent line). Knowing that y'=1/(x-sin(y)x), the bottom of this equation should equal to 0. Thus, set this equation to 0 and solve for y; your answer should be pi/2.
Let f(x) = sin(x^4)
Is the function concave up or concave down at x=10?
f''(x)=12x^2(cos(x^4))+4x^3(4x^3)(-sin(x^4))
f''(x)=4888687.635
The second derivative is positive, therefore the graph of f(x) should be concave up.
Why do none of my jokes work?
Because they're about unemployed people.
Let f be a differentiable function that's
-defined everywhere
-has x=13 as its only x intercept
-has x=1 as a local max which is its only critical point
Let g(x) = (1/f(x^2-3))^(2)+3
How many critical points does g(x) have?
The five critical points found should be +/-4, +/-2, and 0.
The best quadratic approximation to y=2lnx is given in the form displayed below
y=ax^2+bx+c
where a, b, and c are constants. Knowing that this approximation has the same first and second derivatives as y=2lnx at x=1, solve for b.
First, solve for the first and second derivatives of the original function and its approximation, then solve for a. Knowing that a=-1, set each respective function's first derivative to each other, plug in a, and solve for b. Your answer should be 4.
Let the function y(x) be defined implicitly on 0<x<0 by ln(x/e^y)=cos(y).
Find the x coordinate where this curve y(x) has a vertical tangent line for 0<x<10.
Unfortunately, this x coordinate cannot be solve for using the derivative of y(x). However, if we know the y coordinate where y(x) has a tangent line (which can be solve for using y(x)'s derivative), then we can plug this in into the original function to solve for its corresponding x coordinate. By substituting a y coordinate of pi/2, the resulting x coordinate should be e^pi/2.
Let f(x)=(cos(ktan(x)))/(pi)^(sin(2x)-1) and let the equation of the tangent line to y=f(x) at x=pi/4 be represented by the following:
y=-{ }ksin({ }k)(x - pi/{ }) + { }
What value belongs in the third space?
4
How do you make holy water?
You boil the hell out of it.
Let f be a differentiable function that's
-defined everywhere
-has x=13 as its only x intercept
-has x=1 as a local max which is its only critical point
Let g(x) = (1/f(x^2-3))^(2)+3
Classify the critical point at 0.
The best quadratic approximation to y=2lnx is given in the form displayed below
y=ax^2+bx+c
where a, b, and c are constants. Knowing that this approximation has the same first and second derivatives as y=2lnx at x=1, solve for c.
First, solve for a and b by using the first and second derivatives of each respective function. Then, find the corresponding y coordinate to x=1 by substituting 1 as x into the original function. By plugging in this y value, x value, and the a and b value into the linear approximation, c can be solved for, which should equal -3.
Find the rate of change of volume with respect to pressure. The original function is shown below with a, b, n, and R as constants:
(P+(n^2a/V^2))(V-nb)=nRT
First, distribute the entire given function - this should give you PV-Pnb+(n^2a/V^2)-(n^3ab/V^2)=nRT. Next, take the derivative of this function by using the product and quotient rules; after simplifying this derivative, you should get the answer below:
V'=(nb-V)/(P-(n^2a/V^2)+(2n^3ab/V^3)
Let f(x)=(cos(ktan(x)))/(pi)^(sin(2x)-1) and let the equation of the tangent line to y=f(x) at x=pi/4 be represented by the following:
y=-{ }ksin({ }k)(x - pi/{ }) + { }
What value belongs in the fourth space?
Plug in the x value, pi/4, into the original equation to solve for its corresponding y value, which belongs in the fourth space. You should get 3 as your answer.
Why can you never trust a math teacher holding graphing paper?
They must be plotting something.
Let f be a differentiable function that's
-defined everywhere
-has x=13 as its only x intercept
-has x=1 as a local max which is its only critical point
Let g(x) = (1/f(x^2-3))^(2)+3
Classify the critical point at x=-2.
Before using the first derivative test, sketch a graph of f(x) based on the given information. Afterwards, use the first derivative test to find that at x=-2 there is a local maximum.
Find the equation of the tangent line at the x-intercept of the curve y(x), where y(x)=ln(x/e^y)=cos(y).
First, find the coordinates of the x intercept by plugging in 0 for y into the given implicitly defined function. Then, solve for the derivative of its function and plug in 1 into x to find the slope of the tangent line at x=1. The resulting linear approximation should be: y= (1/e)(x-e)+0
Let f(x)=(cos(ktan(x)))/(pi)^(sin(2x)-1) and let the equation of the tangent line to y=f(x) at x=pi/4 be represented by the following:
y=-{ }ksin({ }k)(x - pi/{ }) + { }
What value belongs in the first and second space?
The first and second space are both a part of the slope of the linear approximation, aka the derivative of f(x). To find this value, begin by solving for the derivative of f(x), then plug in pi/4 into that derivative. A slope of -2ksin(k) should be found; thus the first space is 2 whereas the second is 1.
What do you call dudes who love math?
AlgeBROS