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MathNote: If y = x^n ; where n = constant, then dy/dx = n*x^(n1) Now, `y = 1/{(x^2) + 4}` `Thus, y = {(x^2) + 4}^1` `or, y' = 1{{(x^2) + 4}^2}*(2x)` `or, y' = 2x/{(x^2)+4}^2` ``

MathNote : 1) If y = x^n ; where n = constant, then dy/dx = n*(x^(n1)) 2) If y = n*x ; where n = constant ; then dy/dx = n Now, `y = 1/(5x+1)^2` `or, y = (5x+1)^2` `thus, dy/dx = y' =...

MathNote: If y = cos(ax) ; then dy/dx = a*sin(ax) Now, `y = 5cos(9x + 1)` `dy/dx = y' = 5*(sin(9x+1))*9` `or, dy/dx = y' = 45sin(9x+1)` ``

Mathy = 1 cos(2x) + 2`(cos^2x)` `dy/dx = y' = 2*sin(2x)  4*cosx*sinx` `or, y' = 2sin(2x)  2sin(2x) = 0` note: `2sinx*cosx = sin2x`

MathNote: 1) If y = sin(x) ; then dy/dx = cos(x) 2) If y = x^n ; then dy/dx = n*x^(n1) ; where n = constant Now, `y = (x/2)  {sin(2x)/4}` `dy/dx = y' = (1/2)  {2cos(2x)}/4` `or, dy/dx = y' = (1/2)...

MathGiven: `y=(sec(x))^7/7(sec(x))^5/5` `y'=(7sec(x)^6)/7sec(x)tan(x)(5sec(x)^4)/5sec(x)tan(x)` `y'=(sec(x))^7tan(x)(sec(x))^5tan(x)` `y'=(sec(x))^5tan(x)[(sec(x))^21]`...

MathYou need to differentiate the function with respect to x, using the product rule and chain rule, such that: `y' = x'*(6x + 1)^5 + x*((6x + 1)^5)'` `y' = 1*(6x + 1)^5 + x*5*(6x + 1)^4*(6x+1)'` `y' =...

Math`f(s)=(s^21)^(5/2)(s^3+5)` Using the product rule for the derivative, `f'(s)=(s^21)^(5/2)d/(ds)(s^3+5)+(s^3+5)d/(ds)(s^21)^(5/2)` `f'(s)=(s^21)^(5/2)(3s^2)+(s^3+5)(5/2)(s^21)^((5/2)1)(2s)`...

Math`f(x)=3x/sqrt(x^2+1)` Using the quotient rule of the derivative, `f'(x)=3((sqrt(x^2+1)xd/(dx)(sqrt(x^2+1)))/((sqrt(x^2+1))^2))` `f'(x)=3((sqrt(x^2+1)x(1/2)(x^2+1)^(1/2)2x)/(x^2+1))`...

MathGiven: `h(x)=((x+5)/(x^2+3))^2` To find the derivative of the function use the Quotient Rule within the Chain Rule. `h(x)=2((x+5)/(x^2+3))[((x^2+3)(1)(x+5)(2x))/(x^+3)^2]`...

Math`f(x)=sqrt(1x^3)` `f'(x)=(1/2)(1x^3)^(1/21)(3x^2)` `f'(x)=(3x^2)/(2sqrt(1x^3))` Therefore the derivative of the function at the point (2,3) can be obtained by plugging in the value of of x...

Math`f(x)=root(3)(x^21)` ` ` `f'(x)=(1/3)(x^21)^(1/31)(2x)` `f'(x)=(2x)/(3(x^21)^(2/3))` Therefore the derivative at the point(3,2) can be obtained by plugging in the value of x in the f'(x). f'(x)...

MathIn order to find the slope of a function at a given point, find the derivative of the function then plug in the xvalue. Given: `f(x)=3x^24x, (1, 1)` `f'(x)=6x` `f'(1)=6(1)` `f'(1)=6` ``

MathIn order to find the slope of a function at a specific point, first find the derivative of the function, then plug in the xvalue from the given point. Given: `f(x)=2x^48, (0,8)` `f'(x)=8x^3`...

MathIn order to find the slope of the given function at a specified point, first find the derivative of the function, then evaluate the derivative of the function using the given xvalue. Given:...

MathYou need to evaluate the derivative of the given function and since the function is a product of two polynomials, then you must use the product rule, such that: `f'(x) = (5x^2 + 8)'(x^2  4x  6) +...

MathYou need to evaluate the derivative of the given function and since the function is a product of two polynomials, then you must use the product rule, such that: `f'(x) = (2x^3 + 5x)'(3x4) + (2x^3...

MathYou need to evaluate the derivative of the given function and since the function is a product of two polynomials, then you must use the product rule, such that: `f'(x) = (sqrt x)'(sin x) + (sqrt...

MathYou need to evaluate the derivative of the given function and since the function is a product of two polynomials, then you must use the product rule, such that: `f'(t) = (2t^5)'(cos t) + (2t^5)(cos...

MathYou need to evaluate the derivative of the given function and since the function is a quotient of two polynomials, then you must use the quotient rule, such that: `f'(x) = ((x^2+x1)'(x^21) ...

MathYou need to evaluate the derivative of the given function and since the function is a quotient of two polynomials, then you must use the quotient rule, such that: `f'(x) = ((2x+7)'(x^2+4) ...

Math`y=x^4/cosx` Using the quotient rule for evaluating the derivative of the function, `y'=(cosxd/dx(x^4)x^4d/dxcosx)/(cosx)^2` `y'=(cos(x)*(4x^3)x^4(sin(x)))/(cos^2(x))`...

MathYou need to evaluate the derivative of the given function and since the function is a quotient of two functions, then you must use the quotient rule, such that: `f'(x) = ((x^4)(sin x)'  (x^4)'(sin...

MathYou need to evaluate the derivative of the given function, using the product tule for the product `3x^2*sec x` , such that: `f'(x) = (3x^2)'*(sec x) + 3x^2*(sec x)'` `f'(x) = 6x*(sec x)+ 3x^2*(sec...

MathYou need to evaluate the derivative of the given function, using the product tule for the product `x^2*tan x` , such that: `f'(x) = (2x)'  ((x^2)'(tan x) + (x^2)(tan x)') ` `f'(x) = 2  2x*(tan...

MathYou need to evaluate the derivative of the given function, using the product rule for the product x*cos x, such that: `f'(x) = ((x)'(cos x) + (x)(cos x)')  (sin x)'` `f'(x) = 1*cos x + x*(sin x)...

MathYou need to evaluate the derivative of the given function, using the product tule for the products `3x*sin x ` and `x^2*cos x` , such that: `f'(x) = (3x)'*(sin x) + 3x*(sin x)' + (x^2)'(cos x) +...

MathGiven: `f(x)=(x+2)(x^2+5),(1,6)` Find the derivative using the Quotient Rule. Then substitute in the given x value into the f'(x) equation to calculate the slope. `f'(x)=(x+2)(2x)+(x^2+5)(1)`...

MathGiven: `f(x)=(x4)(x^2+6x1),(0, 4)` Find the derivative of the function using the Product Rule. Then plug in the given x value into the derivative function to calculate the slope....

Math`f(x) = (x+1)/(x1)` `f'(x) = [(x1)*1  (x+1)*1]/(x1)^2` `or, f'(x) = (2)/(x1)^2` Now, slope of the tangent at point ((1/2),3) = f'(1/2) = 8 Thus, equation of the tangent to the curve at the...

MathHello! `f(x) = (1+cos(x))/(1cos(x)).` First, let's check that `f(pi/2) = 1` : `cos(pi/2) = 0` and (1+0)/(10) = 1. Then remember that`1+cos(x) = 2*[cos(x/2)]^2` and `1cos(x) = 2*[sin(x/2)]^2` ....

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*x^(n1) Now, `g(t) = 8t^3 5t + 12` `g'(t) = 8*3*t^2  5 + 0` `or, g'(t) = 24t^2  5` `Thus, g''(t) = 24*2*t^1  0` `or, g''(t) = 48t` ``

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*x^(n1) Now, `h(x) = 6x^2 + 7x^2` `h'(x) = 12x^3 + 14x` `h''(x) = 12*(3)*x^4 + 14` `or, h''(x) = 36x^4 + 14` ``

MathYou need to find derivative using limit definition, such that: `f'(x)= lim_(Delta x > 0) (f(x + Delta x)  f(x))/(Delta x)` `f'(x) = lim_(Delta x > 0) (5(x + Delta x)  4  5x + 4)/(Delta...

MathYou need to find derivative using limit definition, such that: `f'(x)= lim_(Delta x > 0) (f(x + Delta x)  f(x))/(Delta x)` `f'(x) = lim_(Delta x > 0) ((x + Delta x)^2  4(x+Delta x) + 5 ...

MathYou need to find derivative using limit definition, such that: `f'(x)= lim_(Delta x > 0) (f(x + Delta x)  f(x))/(Delta x)` `f'(x) = lim_(Delta x > 0) (6/(x+Delta x)  6/x)/(Delta x)`...

MathGiven the function `f(x)=2x^23x` , c=2. We have to use the alternative form of derivative to find the derivative at x=c. Here given that c=2. So by definition of alternative derivative we have,...

MathGiven the function `f(x)=1/(x+4)` We have to use the alternative form of derivative to find the derivative at x=c. Here given that c=3. So by definition of alternative derivative we have,...

MathAs per the rules of differentiation, derivative of a constant is zero Thus, If y = 25 dy/dx = y' = 0

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*x^(n1) Now, ` y = 4t^4` `or, y' = 4*4*t^3` `or, y' = f'(t) = 16*t^3` ``

MathNote: If y = x^n ; then dy/dx = n*x^(n1) ; where n = constant Now, `f(x) = x^3  11x^2` `f'(x) = 3*x^2  11*2*x^1` `or, f'(x) = 3x^2  22x` ``

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*x^(n1) Now, `g(s) = 3s^5  2s^4` `g'(s) = 3*5*s^4  2*4*s^3` `or, g'(s) = 15s^4  8s^3` ``

MathGiven: `h(x)=6sqrt(x)+3root(3)(x)` Rewrite the function as `h(x)=6x^(1/2)+3x^(1/3)` To find the derivative, use the power rule for derivatives. `h'(x)=(1/2)6x^(1/2)+(1/3)3x^(2/3)`...

MathNote: If y = x^n ; then dy/dx = n*x^(n1) ; where n = constant Now, `f(x) = x^(1/2)  x^(1/2)` `f'(x) = (1/2)*x^{(1/2)1}  (1/2)*x^{(1/2)1}` `or, f'(x) = (1/2)*x^(1/2) + (1/2)*x^(3/2)`...

MathGiven: `g(t)=2/(3t^2)` First rewrite the function as `g(t)=(2t^2)/3` Find the derivative of the function by using the power rule. `g'(t)=2(2/3)t^2` `g'(t)=(4/3)t^3` The derivative is...

MathFind the derivative of `h(x)=(8)/(5x^4)` First rewrite the function as `h(x)=(8x^4)/(5)` then find the derivative of the function using the power rule. `h'(x)=(4)(8/5)x^5` The derivative is:...

MathGiven: `f(theta)=4theta5sin(theta)` `g'(theta)=45cos(theta)` ``

MathGiven: `g(alpha)=4cos(alpha)+6` `g'(alpha)=4sin(alpha)` `<br datamcebogus="1">` `<br datamcebogus="1">`

MathGiven: `f(theta)=3cos(theta)(1/4)sin(theta)` The derivative is: `f'(theta)=3sin(theta)(1/4)cos(theta)` ``

MathGiven: `g(alpha)=5(sin(alpha)/3)2alpha` `g'(x)=(5/3)cos(alpha)2` ``