# Homework Help

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• Math
You need to use the chain rule to evaluate the derivative of the function, such that: `y' = (sin(pi*x)^2)'*((pi*x)^2)'*(pi*x)'` `y' = (cos(pi*x)^2)*(2(pi*x))*pi` `y' = 2pi^2*x(cos(pi*x)^2)` Hence,...

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• Math
Given: `y=(4x-1)^3` To find the derivative of the function use the Chain Rule. `y'=3(4x-1)^2(4)` `y'=12(4x-1)^2`

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• Math
Given: `y=5(2-x^3)^4` To find the derivative of the function use the Chain Rule. `y'=20(2-x^3)^3(-3x^2)` `y'=-60x^2(2-x^3)^3`

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• Math
Given: `g(x)=3(4-9x)^4` Find the derivative of the function by using the Chain Rule. `g'(x)=12(4-9x)^3(-9)` `g'(x)=-108(4-9x)^3`

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• Math
`f(t) = (9t+2)^(2/3)` `f'(t) = (2/3)*9*(9t+2)^(-1/3)` `or, f'(t) = 6*(9t+2)^(-1/3)` ` `

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• Math
`f(t) = sqrt(5-t)` `f'(t) = 1/{2sqrt(5-t)}*(-1)` `or, f'(t) = -1/{2sqrt(5-t)}` ` ` ` ` Note:- If y = sqrt(x) ; then dy/dx = 1/{2sqrt(x)}

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• Math
`g(x) = sqrt(4 - 3x^2)` `or, g(x) = (4- 3x^2)^(1/2)` `thus, g'(x) = (1/2)*(4-3x^2)^(-1/2)*(-6x)` `or, g'(x) = -(3x)/(4-3x)^(1/2)` `` ` `

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• Math
Given `y=root(3)(6x^2+1)` Rewrite the function as `y=(6x^2+1)^(1/3)` Find the derivative using the Chain Rule. `y'=(1/3)(6x^2+1)^(-2/3)(12x)` `y'=4x(6x^2+1)^(-2/3)` `y'=(4x)/(6x^2+1)^(2/3)`...

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• Math
Given `f(x)=sqrt(x^2-4x+2)` Rewrite the function as `f(x)=(x^2-4x+2)^(1/2)` Find the derivative using the Chain Rule. `f'(x)=(1/2)(x^2-4x+2)^(-1/2)(2x-4)` `f'(x)=(1/2)(2(x-2))/(x^2-4x+2)^(1/2)`...

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• Math
Given `y=2root(4)(9-x^2)` Rewrite the function as `y=2(9-x^2)^(1/4)` Find the derivative using the Chain Rule. `y'=(1/4)(2)(9-x^2)^(-3/4)(-2x)` `y'=-1x(9-x^2)^(-3/4)` `y'=-x/(9-x^2)^(3/4)`...

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• Math
Given: `f(x)=root(3)(12x-5)` Rewrite the function as `f(x)=(12x-5)^(1/3)` Find the derivative by using the Chain Rule. `f'(x)=(1/3)(12x-5)^(-2/3)(12)` `f'(x)=4(12x-5)^(-2/3)`...

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• Math
Given: `y=1/(x-2)` Rewrite the function as `y=1(x-2)^-1` Find the derivative using the Chain Rule. `y'=-1(x-2)^-2` `y'=(-1)/(x-2)^2`

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• Math
`s(t) = 1/[4-5t-t^2]` `or, s'(t) = -(-5-2t)/[4-5t-t^2]^2` `or, s'(t) = (5+2t)/[4-5t-t^2]^2` ` ` ` `

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• Math
`f(t) = 1/(t-3)^2` `or, f(t) = (t-3)^-2` `Thus, f'(t) = -2(t-3)^-3` `or, f'(t) = -2/(t-3)^3` ``

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• Math
Given: `f'(x)=x^2` `f''(x)=2x` ``

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• Math
Given: `f''(x)=2-(2/x)` Rewrite the function as `f''(x)=2-2x^-1` `f'''(x)=2x^-2` `f'''(x)=2/x^2` ``

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• Math
Given `f'''(x) = 2sqrt(x)` `Thus, f''''(x) = 2/{2sqrt(x)}` `or, f''''(x) = 1/sqrt(x)` ``

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• Math
Given:- `f''''(x) = 2x+1` `Thus, f'''''(x) = 2` `and f''''''(x) = 0` ``

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• Math
The function f(x) = sin x*(sin x + cos x). Use the product rule to determine the derivative of f(x). f'(x) = (sin x)'*(sin x + cos x) + sin x*(sin x + cos x)' f'(x) = (cos x)*(sin x + cos x) +...

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• Math
The function `f(x) = (x^3+4x-1)*(x-2)` . The slope of the tangent at point x = a, is given by f'(a). `f'(x) = ((x^3+4x-1)*(x-2))'` = `(x^3+4x-1)'*(x-2)+ (x^3+4x-1)*(x-2)'` = `(3x^2+4)*(x-2)+...

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• Math
Start by taking the derivative. This will require product rule and chain rule. The product rule is: AB'+BA' `f'(x) = (x-2)(2x) + (x^2+4)(1)` `f'(x)= (x-2)(2x) + x^2+4` `f'(x)= 2x^2 -4x+x^2+4` The...

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• Math
You need to evaluate the equation of the tangent line at (-5,5), using the formula: `f(x) - f(-5) = f'(-5)(x + 5)` Notice that f(-5) = 5. You need to evaluate f'(x), using the quotient rule, such...

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• Math
You need to evaluate the equation of the tangent line at (4,7), using the formula: `f(x) - f(4) = f'(4)(x - 4)` Notice that f(4) = 7. You need to evaluate f'(x), using the quotient rule, such...

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• Math
You need to evaluate the equation of the tangent line to the curve `f(x) = tan x` , t the point `((pi)/4, 1), ` using the following formula, such that: `f(x) - f((pi)/4) = f'((pi)/4)(x - (pi)/4)`...

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• Math
You need to evaluate the equation of the tangent line to the curve` f(x) = sec x` , t the point `((pi)/3, 2), ` using the following formula, such that: `f(x) - f((pi)/3) = f'((pi)/3)(x - (pi)/3)`...

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• Math
Given `f(x)=(2x-1)/(x^2)` Find the derivative of the function using the Quotient Rule. Set the derivative equal to zero to find the critical x value(s). When the derivative is equal to zero the...

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• Math
Given: `f(x)=x^2/(x^2+1)` `` ` ` Find the derivative of the function using the Quotient Rule. Set the derivative equal zero to find the critical x value(s)....

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• Math
Given `f(x)=x^2/(x-1)` Find the derivative of the function using the Quotient Rule. Set the derivative equal to 0 and solve for the x value(s). When the derivative is equal to zero, the slope of...

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• Math
Given: `f(x)=(x-4)/(x^2-7)` Find the derivative of the function using the Quotient Rule. Set the derivative equal to zero and solve for the critical x value(s). When the derivative is zero the...

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• Math
Note:- If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) If y = k ; where k = constant ; then dy/dx = 0 Now, `f(x) = x^4 + 2x^3 - 3x^2 - x` `f'(x) = 4x^3 + 6x^2 - 6x - 1` `f''(x) = 12x^2 +...

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• Math
You need to evaluate the first derivative of the function: `f'(x) = (4x^5-2x^3+5x^2)'` `f'(x) = (4x^5)'-(2x^3)'+(5x^2)'` `f'(x) =20x^4 - 6x^2 + 10x` You need to evaluate the second derivative of...

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• Math
Note:- If x = x^n ; where n = constant ; then dy/dx = n*x^(n-1) Now, `f(x) = 4x^(3/2)` `f'(x) = 4*(3/2)*x^(1/2)` `or, f'(x) = 6*x^(1/2)` `f''(x) = 6*(1/2)*x^(-1/2)` `or, f''(x) = 3*x^(-1/2)` ``

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• Math
You need to evaluate the first derivative of the function, using the quotient rule for `3x^(-3):` `f'(x) = 2x - (9x^2)/(x^6)` You need to simplify by `x^2:` `f'(x) = 2x - (9)/(x^4)` You need to...

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• Math
`f(x) = x/(x-1)` `f'(x) = [(x-1) - x]/(x-1)^2` `or, f'(x) = -1/(x-1)^2` `f''(x) = 2/(x-1)^3` `or, f''(x) = 2(x-1)^-3` ` `

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• Math
`f(x) = (x^2 + 3x)/(x-4)` `f'(x) = [(x-4)(2x+3) - (x^2 + 3x)]/(x-4)^2` `or, f'(x) = [2x^2 + 3x - 8x - 12 - x^2 - 3x]/(x-4)^2` `or, f'(x) = (x^2 - 8x - 12)/(x-4)^2` `Now,` `f''(x) = [(x-4)^2*(2x-8)...

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• Math
`f(x) = xsin(x)` `f'(x) = xcos(x) + sin(x)` `f''(x) = -xsin(x) + cos(x) + cos(x)` `or, f''(x) = 2cos(x) - xsin(x)` ``

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• Math
`f(x) = sec(x)` `f'(x) = sec(x)tan(x)` `f''(x) = sec(x)*sec^2(x) + tan(x)*sec(x)tan(x)` `or, f''(x) = sec(x)[sec^2(x) + tan^2(x)]` ``

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• Math
`f(t) = (t^2)sin(t)` `f'(t) = 2tsin(t) + (t^2)cos(t)` `or, f'(t) = t[2sin(t) + tcos(t)]` ``

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• Math
Note:- if y = cos(x) ; then dy/dx = -sinx Now, ` ` `f(theta) = {(theta)+1}cos(theta)` thus, `f'(theta) = cos(theta) - {(theta) + 1}sin(theta)` `` ` `

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• Math
Note:- 1) If y = cosx ; then dy/dx = -sinx 2) If y = u/v ; where 'u' & 'v' are functions of x or constants ; then dy/dx = [vu'-uv']/(v^2) Now, `f(t) = cost/t` `f'(t) = [-t*sint - cost]/t^2`...

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• Math
Note:- 1) If y = sinx ; then dy/dx = cosx 2) If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) 3) If y = u/v ; where 'u' & 'v' are functions of 'x' ; then dy/dx = [vu' - uv']/(v^2) Now,...

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• Math
Note:- 1) If y = tanx ; then dy/dx = sec^2(x) 2) If y = x^n ; then dy/dx = n*x^(n-1) Now, `f(x) = -x + tanx` `f'(x) = -1 + sec^2(x)` ``

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• Math
Note:- 1) If y = cotx ; then dy/dx = -cosec^2(x) 2) If y = x^n ; where 'n' = constant ; then dy/dx = n*x^(n-1) Now, `y = x+cot(x)` `dy/dx = y' = 1 - cosec^2(x)` `or, dy/dx = -[cosec^2(x) - 1]`...

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• Math
`g(t) = t^(1/4) + 6cosec(t)` `g'(t) = (1/4)t^(-3/4) - 6cosec(t)cot(t)` ` `Note:- 1) If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) 2) If y = cosec(x) ; then dy/dx = -cosec(x)*cot(x)

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• Math
Note:- 1) If y = secx ; then dy/dx = sec(x)*tan(x) 2) If y = x^n ; where n = constant ; then dy/dx = n*x^(n-1) Now, `y = 1/x - 12sec(x)` `y' = dy/dx = -1/x^2 - 12sec(x)*tan(x)` ``

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• Math
You need to evaluate the derivative of the given function, using the quotient rule, such that: `y' = ((3 - 3sin x)'*(2 cos x) - (3 - 3 sin x)*(2 cos x)')/(4 cos^2 x)` `y' = (-3cos x*(2 cos x) + 2...

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• Math
Note:- 1) If y = secx ; then dy/dx = secx*tanx 2) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = constant 3) If y = u/v ; where 'u' & 'v' are functions of 'x' , then dy/dx = [vu'-uv']/(v^2)...

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• Math
Note:- 1) If y = sinx ; then dy/dx = cosx 2) If y = cosecx ; then dy/dx = -cosecx*cotx Now, `y = -cosec(x) - sin(x)` `dy/dx = y' = cosec(x)*cot(x)-cos(x)` ``

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• Math
Note:- 1) If y = sinx ; then dy/dx = cosx 2) If y = cosx ; then dy/dx = -sinx 3) If y = u*v; where 'u' & 'v' are functions of 'x' ; then dy/dx = uv' + vu' Now, `y = xsinx + cosx` `dy/dx = y' =...

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