eNotes Homework Help is a way for educators to help students understand their school work. Our experts are here to answer your toughest academic questions! Once it’s posted to our site, your question could help thousands of other students.
Popular Titles
Showing
in

Math`y = ln[secx + tanx]` `dy/dx = {1/(secx + tanx)}*{secx*tanx + sec^2(x)` ` ` `dy/dx = {secx*(secx + tanx)}/(secx+tanx) = secx` `(d^2y)/dx^2 = secx*tanx` ` `

MathThis is a plot of `f(x) `, we will find the domain of `f(x) `, notice the missing values represented by circles in the plot. Given `f(x) = x/(1ln(x1))` , we know that the denominator...

Math`f(x)=sqrt(2+ln(x)) `. To find the domain of `f(x)` we first need to find out what values does `ln(x)`take. Given below is a plot of `ln(x)` We see that its domain is `x>0` . Since `sqrt(x)`...

MathDifferentiate `f(x)=ln(x^22x) ` and find the domain of f: The domain is all possible inputs. The natural logarithm is defined for all inputs greater than 0, so `x^22x>0 ==> x<0,x>2 `...

Math`f(x)=ln(ln(ln(x))) `. To find the domain of `f(x)` we need to solve the inequality `ln(ln(x))>0 `. Take the exponential on both sides,...

Math`f(x) = lnx/(x^2)` `f'(x) = (1/x^3)  ((2*lnx)/(x^3))` `or, f'(x) = (1/x^3)*(1  2*lnx)` `Now, f'(1) = (1/1)*(12*ln1)` `or, f'(1) = 1` Note: ln1 = 0

Math`f(x) = ln(1 + e^(2x))` `f'(x) = {1/(1+e^(2x))}*(2*e^(2x))` `Now, f'(0) = {1/(1+e^(0))}*(2*e^(0))` `or, f'(0) = (1/2)*2 = 1` Note: `e^0 = 1`

MathThe equation of the tangent line to the curve `y = ln (x^2  3x + 1)` , at the point (3,0) is the following, such that: `f(x)  f(x_0) = f'(x_0)(x  x_0)` You need to put ` ` `f(x) = y, f(x_0) = 0,...

MathThe equation of the tangent line to the curve `y = x^2*ln x` , at the point (1,0) is the following, such that: `f(x)  f(x_0) = f'(x_0)(x  x_0)` You need to put` f(x) = y, f(x_0) = 0, x_0 = 1` You...

MathGiven: `y=(x^2+2)^2(x^4+4)^4` Take the logarithm of both sides of the equation, then rewrite the expression using the laws of logarithms. `` `lny=ln[(x^2+2)^2(x^4+4)^4]`...

Math`y=(e^x*(cos(x))^2)/(x^2+x+1)` Taking the natural logarithm of both sides and applying the properties of logarithms, we get `logy=loge^x+2logcosx log(x^2+x+1)`...

MathGiven: `y=sqrt((x1)/(x^4+1))` Take the logarithm of both sides of the equation. Then rewrite the equation using the Laws of Logarithms. `lny=lnsqrt((x1)/(x^4+1))`...

Math`y=sqrt(x)e^(x^2x) (x+1)^(2/3)` Taking the natural logarithm of both the sides and applying the properties of the logarithms, we get `lny=1/2lnx+(x^2x)+2/3ln(x+1)` Differentiating both sides with...

MathSince variable is raised to a variable power in this function, the common rules of differentiation do not work. First, you need to apply the natural logarithm both sides, such that: `ln y =...

Math`y = x^(cosx)` taking log to the base 'e' both sides we get `lny = (cosx)*lnx` Thus, `(1/y)*dy/dx = sinx*(lnx) + (cosx/x)` `or, dy/dx = y*[(cosx/x)  sinx*(lnx)]` `or, dy/dx = x^(cosx)*[(cosx/x) ...

MathYou need to differentiate with respect to x, the given function, such that: `f'(x) = (xln x  x)'` `f'(x) = (xln x)'  x'` You need to use the product rule to differentiate xln x, such that: `(xln...

MathYou need to differentiate the function with respect to x, using the chain rule, such that: `y' = (sin(ln x))'` `y' = sin'(ln x)*(ln x)'*(x)'` `y' =cos(ln x)*(1/x)*1` `y' = (cos(ln x))/x` Hence,...

MathYou need to differentiate the function with respect to x, using the chain rule, such that: `f'(x) = (ln(sin(x^2)))'` `f'(x) = (ln'(sin(x^2)))*(sin'(x^2))*(x^2)'` `f'(x) = (1/(sin...

MathYou need to differentiate the function with respect to x, using the chain rule, such that: `f'(x) = (ln(1/x))'` `f'(x) = (ln'(1/x))*(1/x)'` `f'(x) = (1/(1/x))*(1/(x^2))` `f'(x) = x/(x^2)`...

MathYou need to differentiate the function with respect to x, using the quotient rule, such that: `f'(x) = (1/(ln x))'` `f'(x) = (1'*(ln x)  1*(ln x)')/(ln^2 x)` `f'(x) = (0*ln x  1/x)/(ln^2 x)`...

MathGiven: `f(x)=log_(10)(x^3+1)` `f'(x)=(1)/((x^3+1)(ln10))3x^2` `f'(x)=(3x^2)/((x^3+1)(ln10))` The derivative is: `f'(x)=(3x^2)/((x^3+1)(ln10))`

MathGiven: `f(x)=log_(5)(xe^x)` `f'(x)=(1)/((xe^x)ln5)(xe^x+e^x(1))` `f'(x)=(e^x(x+1))/((xe^x)(ln5))` `f'(x)=(x+1)/(xln5)` The derivative is: `f'(x)=(x+1)/(xln5)` ``

Math`f(x) = sinx*ln(5x)` `f'(x) = cosx*ln(5x) + sinx/x` Note: If y = sinx ; then dy/dx = cosx If y = lnx ; then dy/dx = 1/x If y = u*v ; where 'u' & 'v' are functions of 'x'; then dy/dx = u*v' +...

Math`f(u) = u/{1 + lnu}` `f'(u) = {(1+lnu)*((du)/(du))  u*[d(1+lnu)/(du)]}/(1+lnu)^2` `or, f'(u) = [(1+lnu)  (u/u)]/(1+lnu)^2` `or, f'(u) = (1 + lnu  1)/(1+lnu)^2` `or, f'(u) = lnu/(1+lnu)^2` `` ` `

MathGiven: `g(x)=ln(xsqrt(x^21))` Rewrite the equation using the Law of exponents. Then find the derivative. `g(x)=lnx+(1)/(2)ln(x^21)` `g'(x)=(1)/(x)+(1)/(2(x^21))(2x)` The derivative is:...

MathIn order to find h'(x) we need to know a few things: The derivative of log(f(x)), and of sqrt(x^21). We need to know the Chain rule. Given h(x)=log(x+sqrt(x^21)), we start by differentiating...

MathGiven: `G(y)=ln((2y+1)^5/sqrt(y^2+1))` Rewrite the equation using the Law of Logarithms. Then find the derivative of the function. `G(y)=5ln(2y+1)(1)/(2)ln(y^2+1)`...

MathYou need to differentiate the given function, with respect to "r" variable, such that: `g'(r) = (r^2*ln(2r+1))'` Using the product rule, yields: `g'(r) = (r^2)'*ln(2r+1) + r^2*(ln(2r+1))'` You need...

MathYou need to differentiate the function with respect to variable s, using the chain rule, such that: `F'(s) = (ln ln s)'` `F'(s) = ln'(ln s)*(ln s)'*(s)'` `F'(s) = 1/(ln s)*(1/s)*1` `F'(s) = 1/(s*ln...

MathNotes: 1) If y = lnx ; then dy/dx = 1/x 2) If y = x^n ; where n = constant , then dy/dx = n*{x^(n1)} 3) If 'y' is a function which contains subfunctions, then the last function is...

MathNotes: 1) If y = lnx ; then dy/dx = 1/x 2) If y = tanx, then dy/dx = sec^2(x) 3) If y = x^n ; where n = constant , then dy/dx = n*{x^(n1)} 3) If 'y' is a function which contains subfunctions,...

MathNotes: 1) If y = lnx ; then dy/dx = 1/x 2) If y = cosx, then dy/dx = sinx 3) If 'y' is a function which contains subfunctions, then the last function is differentiated first,then the second last...

MathThe integral `int (ln x)^2/x dx` has to be determined. This can be done by substitution. Let y = ln x, taking the derivative with respect to x: `dy/dx = 1/x` or `dy = dx/x` Now substitute this in...

MathThe integral `int dx/(5  3x)` has to be determined. This can be done using substitution. Let `y= 5  3x` Differentiating both the sides with respect to x gives: `dy/dx = 3` `=> dx = (dy)/3`...

MathThe integral `int x^2*e^(x^3) dx` has to be determined. Substitute `x^3 = y` `dy/dx = 3*x^2` `=> (dy)/3 = x^2*dx` Substituting this in the original integral gives: `int x^2*e^(x^3) dx` `= int...

MathPlease look at the attached picture down below for the table of values. The graph should look like this:

MathI have attached the table of values below in a picture format. The graph looks like this:

MathThe table of values are attached below. The graph looks something like this:

MathPlease look at the attached image for the table. The graph looks like this:

MathPlease look at the attached image for the table. The graph looks like this:

MathLook at the image attached for the table of values. The graphs looks like this:

MathIf you graph this function on any tool correctly, you should get this.

MathSimply plug the function into a graphing tool, and you should get this graph.

MathIf you correctly plug this function into any graphing tool, you should get this graph.

MathSimply plug the function into a graphing tool, and you should get this graph.

MathIf you plug this into any graphing utility, you will get this.

MathSimply plug the function into a graphing tool, and you should get this graph.

MathSince you already have the same bases, you just need to set the exponents equal to each other. `3x+2=3 ` Solve for `x ` `x=1/3 `

MathSince you already have the same bases, you just need to set the exponents eqal to each other. `2x1=4 ` Solve for `x ` `x=5/2 `

MathSince you already have the same bases, you just need to set the exponents eqal to each other. `x^(2)3=2x ` Subtract `2x ` from both sides `x^22x3=0 ` Solve the quadratic equation `(x3)(x+1)=0...