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MathGiven `y=x/sqrt(x^2+1)` Rewrite the function as `y=x(x^2+1)^(1/2)` Find the derivative using the Product Rule and Chain Rule. `y'=x[(1/2)(x^2+1)^(3/2)]+(x^2+1)^(1/2)(1)`...

MathYou need to use the product and chain rules to evaluate the derivative of the function, such that: `f'(x) = (1/2)(x^2)'(sqrt(16  x^2)) + (1/2)(x^2)(sqrt(16  x^2))'` `f'(x) = x(sqrt(16  x^2)) +...

MathYou need to use the product and chain rules to evaluate the derivative of the function, such that: `f'(x) = (x)'(sqrt(1  x^2)) + (x)(sqrt(1  x^2))'` `f'(x) = (sqrt(1  x^2)) + (x)((2x)/(2sqrt(1...

MathYou need to use the product and chain rules to evaluate the derivative of the function, such that: `f'(x) = (x)'(2x  5)^3 + (x)((2x  5)^3)'` `f'(x) = (2x  5)^3 + 3x((2x  5)^2)*(2x  5)'`...

MathYou need to use the product and chain rules to evaluate the derivative of the function, such that: `f'(x) = (x^2)'(x  2)^4 + (x^2)((x  2)^4)'` `f'(x) = (2x)(x  2)^4 + 4(x^2)((x  2)^3)*(x2)'`...

Math`g(t) = 1/sqrt(t^2  2)` ` ` `or, g(t) = (t^2  2)^(1/2)` `Thus, g'(t) = (1/2)*(2t)*(t^2  2)^(3/2)` `or, g'(t) = t*(t^2  2)^(3/2)` `or, g'(t) = t/(t^2  2)^(3/2)` `` ` `

Math`y = 1/sqrt(3x+5) = (3x+5)^(1/2)` `thus, dy/dx = y' = (1/2)*{(3x+5)^(3/2)}*3` `or, dy/dx = (3/2)*(3x+5)^(3/2)` `or, dy/dx = y' = (3/2)*{1/(3x+5)^(3/2)}` ``

MathGiven: `y=(3/(t2)^4)` Rewrite the equation as `y=3(t2)^4` Find the derivative using the Chain Rule. `y'=12(t2)^5` `y'=12/(t2)^5`

Math`f(t) = 1/(t3)^2` `or, f(t) = (t3)^2` `Thus, f'(t) = 2(t3)^3` `or, f'(t) = 2/(t3)^3` ``

Math`s(t) = 1/[45tt^2]` `or, s'(t) = (52t)/[45tt^2]^2` `or, s'(t) = (5+2t)/[45tt^2]^2` ` ` ` `

MathGiven: `y=1/(x2)` Rewrite the function as `y=1(x2)^1` Find the derivative using the Chain Rule. `y'=1(x2)^2` `y'=(1)/(x2)^2`

MathGiven: `f(x)=root(3)(12x5)` Rewrite the function as `f(x)=(12x5)^(1/3)` Find the derivative by using the Chain Rule. `f'(x)=(1/3)(12x5)^(2/3)(12)` `f'(x)=4(12x5)^(2/3)`...

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

MathGiven `f(x)=sqrt(x^24x+2)` Rewrite the function as `f(x)=(x^24x+2)^(1/2)` Find the derivative using the Chain Rule. `f'(x)=(1/2)(x^24x+2)^(1/2)(2x4)` `f'(x)=(1/2)(2(x2))/(x^24x+2)^(1/2)`...

MathGiven `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)`...

Math`g(x) = sqrt(43x^2)` `or, g(x) = (43x^2)^(1/2)` `Thus, g'(x) = (1/2){(43x^2)^(1/2)}*(6x)` `or, g'(x) = (3x/2)*(43x^2)^(1/2)` `or, g'(x) = (3x/2)*1/(43x)^(1/2)` `` ` ` ` `

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

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

MathGiven: `g(x)=3(49x)^4` Find the derivative of the function by using the Chain Rule. `g'(x)=12(49x)^3(9)` `g'(x)=108(49x)^3`

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

MathGiven: `y=(4x1)^3` To find the derivative of the function use the Chain Rule. `y'=3(4x1)^2(4)` `y'=12(4x1)^2`

MathGiven: `f''''(x) = 2x+1` `Thus, f'''''(x) = 2` `and f''''''(x) = 0` ``

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

MathGiven: `f''(x)=2(2/x)` Rewrite the function as `f''(x)=22x^1` `f'''(x)=2x^2` `f'''(x)=2/x^2` ``

MathGiven: `f'(x)=x^2` `f''(x)=2x` ``

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)]` ``

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)` ``

Math`f(x) = (x^2 + 3x)/(x4)` `f'(x) = [(x4)(2x+3)  (x^2 + 3x)]/(x4)^2` `or, f'(x) = [2x^2 + 3x  8x  12  x^2  3x]/(x4)^2` `or, f'(x) = (x^2  8x  12)/(x4)^2` `Now,` `f''(x) = [(x4)^2*(2x8)...

Math`f(x) = x/(x1)` `f'(x) = [(x1)  x]/(x1)^2` `or, f'(x) = 1/(x1)^2` `f''(x) = 2/(x1)^3` `or, f''(x) = 2(x1)^3` ` `

MathYou 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...

MathNote: If x = x^n ; where n = constant ; then dy/dx = n*x^(n1) 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)` ``

MathYou need to evaluate the first derivative of the function: `f'(x) = (4x^52x^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...

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*x^(n1) 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 +...

MathGiven: `f(x)=(x4)/(x^27)` 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...

MathGiven `f(x)=x^2/(x1)` 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...

MathGiven: `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)....

MathGiven `f(x)=(2x1)/(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...

MathYou 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)`...

MathYou 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)`...

MathYou 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...

MathYou 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...

MathYou need to evaluate the equation of the tangent line to the curve `f(t) =(sec t)/t` , at the point `(pi, 1/(pi)), ` using the following formula, such that: `f(t)  f(pi) = f'(pi)(t  pi)` Notice...

MathYou need to evaluate the equation of the tangent line to the curve f(x) = tan x*cot x, t the point (1, 1), using the following formula, such that: `f(x)  f(1) = f'(1)(x  1)` Notice that f(`1` ) =...

MathYou need to evaluate the equation of the tangent line to the curve `f(x) =(1 + csec x)/t(1  csc x), ` at the point` (pi/6, 3), ` using the following formula, such that: `f(x)  f(pi/6) =...

Math`h(theta) = 5(theta)sec(theta) + thetatan(theta)` `h'(theta) = 5sec(theta) + 5thetasec(theta)tan(theta) + tan(theta) + thetasec^2(theta)` `or, h'(theta) = thetasec(theta)[sec(theta)+5tan(theta)] +...

Math`y = 2xsin(x) + (x^2)cos(x)` `y' = dy/dx = 2sin(x) + 2xcos(x) + 2xcos(x)  (x^2)sin(x)` `y' = dy/dx = 2sin(x) + 4xcos(x)  (x^2)sin(x)` ` `

Math`f(x) = sin(x)cos(x) = (1/2)sin(2x)` `f'(x) = 2*(1/2)*cos(2x) = cos(2x)` ` `Note: 2sin(x)cos(x) = sin(2x)

MathNote: 1) If y = tanx ; then dy/dx = sec^2(x) 2) If y = x^n ; where 'n' = constant ; then dy/dx = n*x^(n1) 3) If y = u*v ; where 'u' & 'v' are functions of 'x', then dy/dx = uv' + vu' Now,...

MathNote: 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' =...

MathNote: 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)` ``