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MathHello! Consider this function for t in [1, 3] and [3, 5] separately. At the first interval t<=3 and t3 = (t3) = 3t.So y(t) = 3  (3t) = t.The maximum is at the maximum t, i.e. at t=3,...

MathHello! For x<=4 x+4<=0 and g(x) = (x+4).The minimum on the interval [7, 4] is at the x=4, g(4) = 0.The maximum on the interval [7, 4] is at the x=7, g(7) = 3. For x>=4...

Mathf(x) = [x], is the nearest integer function. For values of x in the closed set [2, 2], the maximum value of f(x) is 2 and the minimum value of f(x) is 2. The absolute extrema of the function are...

MathThe function h(x) = [2x] is the nearest integer function. For values of x lying in the closed set [2,2], the maximum value of h(x) = 2  (2) = 2 + 2 = 4. The minimum value of x is 2  2 = 0 The...

MathGiven the function: `f(x)=sin(x)` in the closed interval [5 pi/6 , 11 pi/6] We have to find the absolute extrema of the function on the closed interval. Step 1: Check whether f(x) is a continuous...

MathGiven the function g(x)=sec(x) in the interval [pi/6, pi/3] We have to find the absolute extrema of the function on the closed interval. So first let us find the derivative of the function and...

MathGiven: `y=3cos(x),[0,2pi]` First find the critical x values of the function. To find the critical values, set the derivative equal to zero and solve for the x value(s). `f'(x)=3sin(x)=0`...

MathGiven the function `y=tan((pi x)/8)` in [0,2]. We have to find the absolute extrema of the function in the closed interval [0,2]. So first we have to take the derivative of the function and equate...

MathYou need to evaluate the critical numbers of the function and for this reason, you must differentiate the function with respect to x, such that: `f'(x) = (x^3  3x^2)'` `f'(x) = 3x^2  6x` You need...

MathYou need to evaluate the critical numbers of the function and for this reason, you must differentiate the function with respect to x, such that: `f'(x) = (x^4  8x^2)'` `f'(x) = 4x^3  16x` You...

MathYou need to evaluate the critical numbers of the function and for this reason, you must differentiate the function with respect to t, using the product and chain rules, such that: `g'(t) =...

MathGiven `f(x)=(4x)/(x^2+1)` Find the derivative using the Quotient Rule. The set the derivative equal to zero and solve for the critical x value(s). `f'(x)=[(x^2+1)(4)(4x)(2x)]/(x^2+1)=0`...

MathGiven the function `h(x)=sin^2(x)+cos(x)` in the interval `0<x<2pi` We have to find the critical numbers of the function. First take the derivative of the function and equate it to zero. We...

MathGiven the function f(x) =2sec(`theta)`+tan(`theta)` in the interval 0<theta<2 pi We have to find the critical points. So let us first take the derivative of the function and evaluate it to...

MathGiven: f(x)=3x,[1,2] First find the critical values of the function. To find the critical values of the function (if one exists), set the derivative of the function equal to zero and solve for...

MathGiven: `f(x)=(3/4)x+2,[0, 4]` First find the critical value(s) of the function (if any exist). To find the critical value(s) of the function, set the derivative equal to zero and solve for the...

MathGiven: `g(x)=2x^28x,[0,6]` First find the critical x value(s) of the function. You will do this by setting the derivative function equal to zero and solving for the x value(s). `g'(x)=4x8=0`...

MathGiven: `h(x)=5x^2, [3, 1]` Find the critical values of the function by setting the derivative equal to zero and solving for x. When the derivative is equal to zero the slope of the tangent line...

MathGiven: `f(x)=x^3(3/2)x^2,[1,2]` First find the critical x value(s) of the function. To find the critical x value(s), set the derivative equal to zero and solve for the x value(s)....

Math`f(x) = 4/(x^2 + 1)` `or, f(x) = 4*(x^2 + 1)^1` `thus, f'(x) = 4*(2x)*(x^2 +1)^2` `or, f'(x) = 8x/(x^2 +1)^2` `or, f'(1) = (8*1)/{(1)^2 + 1}^2` `or, f'(x) = 8/4 = 2` ``

Math`f(x) = (3x+1)/(4x3)` `f'(x) = [3*(4x3)  4*(3x+1)]/(4x3)^2` `or, f'(x) = 13/(4x3)^2` `or, f'(4) = 13/(4*4  3)^2` `or, f'(4) = 13/(13)^2` `or, f'(4) = 1/13` ``

Math`y = (1/2)*cosec(2x)` `y' = (1/2)*2*cosec(2x)*cot(2x)` Putting x = pi/4 `y' = 1*cosec(pi/2)*cot(pi/2) = 1*1*0 = 0` ``

Math`y = cosec(3x) + cot(3x)` `y' = 3*cosec(3x)*cot(3x)  3*cosec^2(3x)` Putting x = pi/6 we get y' = `3*cosex(pi/2)*cot(pi/2)  3*cosec^2(pi/2)` `or, y' = 3*1*0  3*1` `or, y' = 3` ``

Math`y = (8x+5)^3` `y' = 3*8*(8x+5)^2` `y' = 24*(8x+5)^2` `y'' = 24*2*8*(8x+5)` `or, y'' = 384*(8x+5)` `` ` `

Math`y = 1/(5x+1) = (5x+1)^1` ` ` `differentiating` `y' = 1{(5x+1)^2}*5` `or, y' = 5(5x+1)^2` ``Differentiating againg w.r.t 'x' we get `y'' = 10*5*(5x+1)^3` `or, y'' = 50/(5x+1)^3` ``

MathNote: 1) If y = cotx ; then dy/dx = `cosec^2(x)` ` ` 2) If y = cosecx ; then dy/dx = cosecx*cotx Now, `f(x) = y = cotx` `differentiating ` `f'(x) = y' = cosec^2(x)` `differentiating` `f''(x) =...

MathNote: If y = sinx ; then dy/dx = cosx If y = cosx ; then dy/dx = sinx 2sinx*cosx = sin(2x) Now, `y = (sinx)^2` `or, y = sin^2x` `differentiating` `y' = 2sinx*cosx` `or,y' = sin(2x)`...

Math`(x^2) + (y^2) = 64` `Differentiating` `2x + 2y(dy/dx) = 0` `or, x + y(dy/dx) = 0` `or, dy/dx = x/y` ``

Math`(x^2) + 4xy  (y^3) = 6` `differentiating` `2x + 4y + 4x(dy/dx)  3(y^2)*(dy/dx) = 0` `or, 2(x+2y) = (dy/dx)*[3(y^2)  4x]` `or, dy/dx = [2(x+2y)]/[3(y^2)4x]` ``

Math`(x^3)*y  x*(y^3) = 4` `differentiating ` `3(x^2)*y + (x^3)*(dy/dx)  (y^3)  3x(y^2)*(dy/dx) = 0` `or, (dy/dx)*[(x^3)3x(y^2)] = (y^3)  3(x^2)y` `or, dy/dx = [(y^3)  3(x^2)y]/[(x^3)3x(y^2)]` ``

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

MathNote: 1) If y = sinx; then dy/dx = cosx 2) If y = cos(x) ; then dy/dx = sinx Now, `x*sin(y) = y*cos(x)` `or, x*cosy*(dy/dx) + sin(y) = y*sin(x) + cos(x)*(dy/dx)` `or, x*cos(y)*(dy/dx) ...

MathNote: If y = cos(x) ; then dy/dx = sin(x) Now, `cos(x+y) = x` `or, sin(x+y)*{1 + (dy/dx)} = 1` `or, 1+(dy/dx) = 1/sin(x+y)` `or, 1+(dy/dx) = cosec(x+y)` `or, dy/dx = 1  cosec(x+y)` `or,...

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*{x^(n1)} Now, `y = 15*x^(5/2)` `y' = 15*(5/2)*x^{(5/2)1}` `or, y' = (75/2)*x^(3/2)` `thus, y'' = (75/2)*(3/2)*x^{(3/2)1}` `or, y'' =...

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*{x^(n1)} Now, `y = 20*x^(1/5)` `y' = 20*(1/5)*x^{(1/5)1}` `or, y' = 4*x^(4/5)` `thus, y'' = 4*(4/5)*x^{(4/5)1}` `or, y'' =...

MathNote: If y = tanx ; then dy/dx = sec^2(x) If y = sec(x) ; then dy/dx = sec(x)*tan(x) Now, `f(theta) = 3tan(theta)` `f'(theta) = 3sec^2(theta)` `f''(theta) = 3*2sec(theta)*sec(theta)*tan(theta)`...

MathNote: If y = cos(ax) ; then dy/dx = a*sin(ax) If y = sin(ax) ; then dy/dx = a*cos(ax) ; where 'a' = constant Now, `h(t) = 10cos(t)  15sin(t)` `h'(t) = 10sin(t)  15cos(t)` `h''(t) = 10cos(t)...

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*{x^(n1)} Now, y = `(7x +3)^4` `thus, dy/dx = y' = 4{(7x+3)^3}*7` `or, dy/dx = y' = 28(7x+3)^3` ``

MathNote: If y = x^n ; where n = constant ; then dy/dx = n*{x^(n1)} Now, `y = (x^2  6)^3` `thus, dy/dx = y' = 3*{(x^2  6)^2}*(2x)` `or, dy/dx = y' = 6x*(x^2  6)^2` ``

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...