# Math Homework Help

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• Math
To find the derivative of the function `f(x)=-10x` using the limit process use the fact that the slope is `m=lim_(h->0)(f(x+h)-f(x))/(h)` `=lim_(h->0)(-10(x+h)-(-10x))/(h)`...

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• Math
By limit process, the derivative of a function f(x) is :- `f'(x) = lim_(h -> 0) [{f(x+h) - f(x)}/h]` Now, the given function is :- `f(x) = 7x -3` THus, `f'(x) =lim_(h -> 0) [{f(x+h) -...

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• Math
The domain of a function y = f(x) is the set of all real values that x an take on for which y is real. If the value of x lies in the domain it is possible to plot the point (x, f(x)). For the...

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• Math
The function f(x) = x^3. To determine f(a+h) substitute x with a+h. f(a+h) = (a+h)^3 = a^3 +3*a^2*h +3*a*h^2+ h^3 f(a) = a^3 `(f(a+h) - f(a))/h` = `(a^3+3*a^2*h +3*a*h^2+ h^3 - a^3)/h` = `(3*a^2*h...

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• Math
Given f(x)=4+3x-x^2, then f(x+h)=4+3(x+h)-(x+h)^2. Consider the difference f(x+h)-f(x): -(x+h)^2+x^2 + 3x+3h-3x + 4-4 = -x^2-2xh-h^2+x^2 + 3h = (3-2x)*h - h^2. This divided by h is equal to...

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• Math
The vertical asymptotes of a curve are lines that the graph of the curve approaches but does not touch. For `y = (f(x))/(g(x))` , the vertical asymptotes are lines x = a where a is the root of the...

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• Math
`d/(dt) sin^-1t=1/sqrt(1-t^2)` `y'=1/(sqrt(1-(2x+1)^2))*d/(dx) (2x+1)` `y'=2/sqrt(1-(4x^2+1+4x))` `y'=2/sqrt(1-4x^2-1-4x)` `y'=2/sqrt(-4x^2-4x)` `y'=2/sqrt(4(-x^2-x))` `y'=2/(2(sqrt(-x(x+1))))`...

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• Math
Find `g'(x) ` if `g(x)=sqrt(x^2-1)sec^(-1)(x) ` : Use the product rule: `g'(x)=sqrt(x^2-1)*1/(|x|sqrt(x^2-1))+(1/2)(x^2-1)^(-1/2)(2x)sec^(-1)(x) ` `=1/|x|+(xsec^(-1)(x))/sqrt(x^2-1) ` or...

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• Math
`d/(dt) cos^-1(t)=(-1)/sqrt(1-t^2)` `G(x)=sqrt(1-x^2)cos^-1(x)` `G'(x)=sqrt(1-x^2) d/(dx) cos^-1(x) + cos^-1(x) d/(dx)sqrt(1-x^2)` `G'(x)=sqrt(1-x^2)*(-1/sqrt(1-x^2)) + cos^-1(x)...

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• Math
`y=tan^-1(x-sqrt(1+x^2))` `d/(dt) tan^-1t=1/(1+t^2)` `y'=1/(1+(x-sqrt(1+x^2))^2) * d/(dx)(x-sqrt(1+x^2))` `y'=1/(1+x^2+1+x^2-2xsqrt(1+x^2)) *(1-(1/2)(1+x^2)^(-1/2)(2x))`...

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• Math
`d/(dx) cot^-1x=(-1)/(1+x^2)` `h(t)=cot^-1(t) + cot^-1(1/t)` `h'(t)=(-1)/(1+t^2) + (-1)/(1+(1/t)^2) *d/(dt) (1/t)` `h'(t)=(-1)/(1+t^2) + (-1)/(1+(1/t)^2) *(-1t^-2)` `h'(t)=(-1)/(1+t^2)...

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• Math
`d/(dx)sin^-1(x)=1/sqrt(1-x^2)` `F(theta)=sin^-1(sqrt(sin(theta)))` `F'(theta)=(1/sqrt(1-sintheta)) * d/(d theta)sqrt(sintheta)` `F'(theta)=(1/sqrt(1-sintheta)) *(1/2)(sintheta)^(-1/2) costheta`...

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• Math
`y=xsin^-1(x) +sqrt(1-x^2)` `y'=xd/(dx) sin^-1(x) + sin^-1(x) d/(dx) x + d/(dx)sqrt(1-x^2)` `y'=x(1/sqrt(1-x^2)) + sin^-1(x) + (1/2)(1-x^2)^(-1/2)(-2x)` `y'=x/sqrt(1-x^2) + sin^-1(x)...

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• Math
`d/(dx)cos^-1(x)=-1/sqrt(1-x^2)` using above `y=cos^-1(sin^-1(t))` `y'=(-1)/sqrt(1-(sin^-1(t))^2) * d/(dt) sin^-1(t)` `y'=(-1)/sqrt(1-(sin^-1(t))^2) * (1/sqrt(1-t^2))`...

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• Math
`y=cos^-1((b+acos(x))/(a+bcos(x)))` `y'=((-1)/sqrt(1-((b+acos(x))/(a+bcos(x)))^2))* d/(dx) ((b+acos(x))/(a+bcos(x)))` `y'=-(a+bcos(x))/sqrt((a+bcos(x))^2-(b+acos(x))^2) *d/(dx)...

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• Math
`d/(dt) tan^-1(t)=1/(1+t^2)` `y=tan^-1(sqrt((1-x)/(1+x)))` `y'=(1)/(1+((1-x)/(1+x))) * d/(dx)sqrt((1-x)/(1+x))` `y'=((1+x)/(1+x+1-x)) * d/(dx) (1-x)^(1/2) (1+x)^(-1/2)` `y'=((1+x)/2) * ((1-x)^(1/2)...

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• Math
Note:- 1) If y = cosx ; then dy/dx = -sinx 2) If y = k ; where k = constant ; then dy/dx = 0 3) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) 4) If y...

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• Math
Note:- 1) If y = sinx ; then dy/dx = cosx 2) If y = e^x ; then dy/dx = e^x 3) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) 4) If y = x^n ; where 'k'...

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• Math
Find `(dy)/(dx) ` if `e^(x/y)=x-y ` by implicit differentiation: Note that if u is a differentiable function of x then `d/(dx)e^u=e^u (du)/(dx) ` . Then: `d/(dx) e^(x/y)=d/(dx)(x-y) ` `e^(x/y)...

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• Math
`sqrt(x+y)=1+x^2y^2` `dy/dxsqrt(x+y)=dy/dx(1+x^2y^2)` `1/2sqrt(x+y)^(-1/2)(1+dy/dx)=x^2(2y)dy/dx+y^2(2x)` `1/(2sqrt(x+y))+(dy/dx)/(2sqrt(x+y))=2x^2ydy/dx+2xy^2`...

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• Math
`tan^-1(x^2y)=x+xy^2` taking derivative on both the sides `d/(dx) tan^-1(x^2y)=d/(dx) (x+xy^2)` `1/(1+(x^2y)^2) * d/(dx) (x^2y)=1+x(2y)dy/dx +y^2` `1/(1+x^4y^2) *(x^2 dy/dx+y(2x)) =1+2xy dy/dx...

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• Math
Differentiate both sides of the equation with respect to x. `d/dx(xsin(y)) + d/dx(ysin(x)) = d/dx 1` Each of these terms needs to be differentiated with the product rule. `x d/dx(siny) + sin(y)d/dx...

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• Math
Note:- 1) If y = cosx ; then dy/dx = -sinx 2) If y = e^x ; then dy/dx = e^x 3) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) 4) If y = sinx ; then...

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• Math
`d/(dx) tan(x-y) =d/(dx)(y/(1+x^2))` `sec^2(x-y) *d/(dx)(x-y) = d/(dx) y(1+x^2)^-1` `(1-dy/dx)sec^2(x-y) = y(-1)(1+x^2)^(-2)(2x) +(1+x^2)^-1 dy/dx` `(1-dy/dx)sec^2(x-y) = (-2xy)/(1+x^2)^2 +...

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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 both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) 4) If y = k ; where...

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• Math
Note:- 1) If y = sinx ; then dy/dx = cosx ; 2) If y = u + v ; where both u & v are functions of 'x' , then dy/dx = (du/dx) + (dv/dx) 3) If y = k ; where 'k' = constant ; then dy/dx = 0 4)...

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• Math
Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = real number 2) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = v*(du/dx) + u*(dv/dx) 3) If y = k ; where 'k' =...

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• Math
Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = real number 2) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) 3) If y = k ; where 'k' =...

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• Math
Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = real number 2) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) 3) If y = k ; where 'k' =...

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• Math
Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = real number 2) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) 3) If y = k ; where 'k' =...

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• Math
Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = real number 2) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) 3) If y = k ; where 'k' =...

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• Math
Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = real number 2) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) 3) If y = k ; where 'k' =...

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• Math
Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = real number 2) If y = tan^(-1)x ; the dy/dx = 1/{1 + (x^2)} Now, the given function contains sub-functions, so the chain rule of...

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• Math
Note:- 1) If y = tan^(-1)x ; then dy/dx = 1/{1+(x^2)} 2) If y = x^n ; then dy/dx = n*x^(n-1) ; where n = real number Clearly, the given function contains sub-functions, so the chain rule of...

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• Math
`x^3+y^3=1` differentiating with respect to x. `3x^2+3y^2(dy/dx)=0` `dy/dx=-x^2/y^2`

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• Math
`2sqrt(x)+sqrt(y)=3` differentiating with respect to x. `2(1/(2sqrt(x)))+(1/(2sqrt(y)))(dy/dx)=0` `dy/dx=(-(2sqrt(y))/sqrt(x))`

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• Math
`x^2+xy-y^2=4` Differentiating with respect to x. We get `2x+(y+x(dy/dx))-2y(dy/dx)=0` `(2x+y)+(x-2y)(dy/dx)=0` `dy/dx=(2y-x)/(2x+y)`

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• Math
Note:- 1) If y = x^n ; then dy/dx = n*x^(n-1) ; where 'n' = real number 2) If y = u*v ; where both u & v are functions of 'x' ; then dy/dx = u*(dv/dx) + v*(du/dx) 3) If y = k ; where 'k' =...

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• Math
`x^4(x+y)=y^2(3x-y)` `x^5+x^4y=3xy^2-y^3` Differentiating with respect to x.We get `5x^4+(4x^3y+x^4(dy/dx))=(3y^2+3x(2y)(dy/dx))-3y^2(dy/dx)` `(5x^4+4x^3y-3y^2)=(6xy-3y^2-x^4)(dy/dx)`...

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• Math
Note:- 1) If y = e^x ; then dy/dx = e^x 2) If y = x^n ; then dy/dx = n*x^(n-1) ; where 'n' = real number 3) If y = u*v ; where both u & v are functions of 'x' ; then dy/dx = u*(dv/dx) +...

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• Math
Note:- 1) If y = cosx ; then dy/dx = -sinx 2) If y = x^2 ; then dy/dx = 2x 3) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) Now, the given function is...

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• Math
Note:- 1) If y = cosx ; then dy/dx = -sinx 2) If y = sinx ; then dy/dx = cosx 3) If y = u*v ; where both u & v are functions of 'x' , then dy/dx = u*(dv/dx) + v*(du/dx) 4) If y = k ; where 'k'...

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• Math
No, sin^2 (x) is the square of sin(x), or sin(x) taken to the second power. By definition, this means sin(x) multiplied by itself two times: `sin^2(x) = sin(x)*sin(x)` . For example, if x = 30...

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• Math
Slope of a tangent line to a function at a point is equal to the derivative of the function at that point. `y=(1+x^3)^(1/2)` `dy/dx=(1/2)*(1+x^3)^((1/2)-1)*3x^2` `dy/dx=(3x^2)/(2(sqrt(1+x^3)))`...

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• Math
`y = pi - x` is the exact answer. At `x = pi` `` `y = sin(sin(pi)) = sin(0) = 0` `y' = 1/2 (cos(x - sin(x)) + cos(x + sin(x))) = 1/2(cos(pi) + cos (pi))` =-1. So y = -x + pi, using the slope...

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• Math
y = x `y' = cos(x) + 2sin(x)cos(x).` x = 0 `y' = 1 + 0 = 1` Substituting into the y-intercept form of the line, `y = 1x + 0 = x`

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• Math
Let `u = sin(pi*x)` `y = 2^(u), (dy)/(du) = 2^(u)log2` `(du)/(dx) = cos(pi*x) * pi` and `y' = pi*log(2)*2^(sin(pi*x))*cos(pi*x)`

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• Math
`y=x^2e^(-1/x)` Derivative can be found by using the product rule `y'= x^2*(e^(-1/x)(-1*-1*x^-2)) +e^(-1/x)*2x` `y'= x^2(x^-2e^(-1/x) )+ 2xe^(-1/x)` `y' = e^(-1/x) +2xe^(-1/x)` `y'=(1+2x)e^(-1/x)`

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• Math
`d/(dy) [cos((1-e^(2x))/(1+e^(2x)))] = -sin((1-e^(2x))/(1+e^(2x)))[d/(dx) ((1-e^(2x))/(1+e^(2x)))]` ` ` `= -sin((1-e^(2x))/(1+e^(2x))){((1+e^(2x))(d/dx)(1-e^(2x)) -...

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