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• 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` ` `

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• Math
This 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/(1-ln(x-1))` , we know that the denominator...

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• 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)`...

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• Math
Differentiate `f(x)=ln(x^2-2x) ` 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^2-2x>0 ==> x<0,x>2 `...

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• 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,...

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• 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)*(1-2*ln1)` `or, f'(1) = 1` Note:- ln1 = 0

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• 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`

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• Math
The 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,...

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• Math
The 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...

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• Math
Given: `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]`...

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• 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)`...

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

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• Math
`y=sqrt(x)e^(x^2-x) (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^2-x)+2/3ln(x+1)` Differentiating both sides with...

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• Math
Since 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 =...

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• 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) -...

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• Math
You 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...

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• Math
You 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,...

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• Math
You 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...

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• Math
You 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)`...

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• Math
You 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)`...

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• Math
Given: `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))`

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• Math
Given: `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)` ``

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• 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' +...

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• 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` `` ` `

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

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• Math
In order to find h'(x) we need to know a few things: The derivative of log(f(x)), and of sqrt(x^2-1). We need to know the Chain rule. Given h(x)=log(x+sqrt(x^2-1)), we start by differentiating...

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• Math
Given: `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)`...

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• Math
You 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...

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• Math
You 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...

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• Math
Notes:- 1) If y = lnx ; then dy/dx = 1/x 2) If y = x^n ; where n = constant , then dy/dx = n*{x^(n-1)} 3) If 'y' is a function which contains sub-functions, then the last function is...

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• Math
Notes:- 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^(n-1)} 3) If 'y' is a function which contains sub-functions,...

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• Math
Notes:- 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 sub-functions, then the last function is differentiated first,then the second last...

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• Math
The 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...

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• Math
The 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`...

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• Math
The 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...

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

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1 TA answer.

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

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1 TA answer.

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

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• Math
Please look at the attached image for the table. The graph looks like this:

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• Math
Please look at the attached image for the table. The graph looks like this:

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• Math
Look at the image attached for the table of values. The graphs looks like this:

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• Math
If you graph this function on any tool correctly, you should get this.

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• Math
Simply plug the function into a graphing tool, and you should get this graph.

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• Math
If you correctly plug this function into any graphing tool, you should get this graph.

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• Math
Simply plug the function into a graphing tool, and you should get this graph.

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• Math
If you plug this into any graphing utility, you will get this.

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• Math
Simply plug the function into a graphing tool, and you should get this graph.

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• Math
Since 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 `

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• Math
Since you already have the same bases, you just need to set the exponents eqal to each other. `2x-1=4 ` Solve for `x ` `x=5/2 `

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• Math
Since 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^2-2x-3=0 ` Solve the quadratic equation `(x-3)(x+1)=0...

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