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If y equals the integral from 5x on top to cosx on bottom of cos of u squared du what is y'?
y=S^5x _cos(x) cos(u²) du The derivative of a definite integral of a function f(x) is equal to the difference in the product of the function at each limit of integration times the limit of integration. y'=cos(u²)*du/dx from u=cos(x) to u=5x y'=-sin(x)*cos(cos(x)²)-5*cos(25x²) To understand why this works, consider the following where F(x) is the antiderivative of f(x) y=F(g(x))-F(h(x))=S f(x)dx from h(x) to g(x) If you take the derivative of this expression and apply the chain rule dy/dx = dF(g(x))/dx - dF(h(x))/dx = f(g(x))*dg/dx - f(h(x))*dh/dx
How do you solve a slope?
You take the change in Y or dy and divide it by the change in X or dx. Slope equals dy/dx.
Integral of sin square root x?
For ∫ sin(√x) dx let y = √x = x1/2 → dy = 1/2 x-1/2 dx → 2x1/2 dy = dx → 2y dy = dx → ∫ sin(x1/2) dx = ∫(sin y) 2y dy Now: ∫ uv dx = u∫v dx - ∫(u'∫v dx) dx → ∫(sin y) 2y dy = ∫2y sin y dy = 2y ∫sin y dy - ∫(2 ∫sin y dy) dy = -2y cos y + 2 sin y + C = 2 sin y - 2y cos y + C → ∫ sin(√x) dx = 2 sin(√x) - 2(√x) cos(√x) + C
What is the derivative of the square root x?
Use: √x = x1/2 By the Power Rule (Decrease the power by 1. Multiply by the original power.): d/dx √x = d/dx x1/2 = 1/2 x-1/2
How do you find the delta x and delta y on a graph?
The delta x and delta y (dx and dy) are the changes in x and in y. When you take the derivative you are provided with a ratio (rise over run) of these changes. Take f(x)=3x^2 + 2x Derivative: (dy/dx)=6x + 2 From this, you can solve for dx and dy algebraically.
How do you find the rate of change for 5x plus 3y equals -2?
Rearrange for y, let f(x) = y, and find the derivative of f(x) with respect to x: 5x + 3y = -2 y = -(5x+2)/3 = f(x) df/dx = -5/3
What is the derivative of y equals -3xsinx - 1.5x to the 2 plus 5x when x equals pi?
y=-3x*sinx-1.5x2+5x, when x=πy'=d/dx(-3x*sinx)-d/dx(1.5x2)+d/dx(5x)y'=(-3x*d/dx(sinx)+sinx*d/dx(-3x))-d/dx(1.5x2)+d/dx(5x)y'=(-3x*cosx+sinx(-3))-d/dx(1.5x2)+d/dx(5x)y'=(-3x*cosx-3sinx)-3x+5y'=-3x*cosx-3sinx-3x+5 is the derivative at any point of that equation, now you only have to plug in π for xy'(π)=-3π*cosπ-3sinπ-3π+5y'(π)=-3π*(-1)-3(0)-3π+5y'(π)=3π-3π+5y'(π)=5
How do you find the derivative of 9 to the 5x?
95x?d/dx(au)=au*ln(a)*d/dx(u)d/dx(95x)=95x*ln(9)*d/dx(5x)-The derivative of 5x is:d/dx(cu)=c*du/dx where c is a constantd/dx(5x)=5*d/dx(x)d/dx(95x)=95x*ln(9)*(5*d/dx(x))-The derivative of x is:d/dx(x)=1x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(95x)=95x*ln(9)*(5*1)d/dx(95x)=95x*ln(9)*(5)-95x can simplify to (95)x, which equals 59049x.-ln(9) can simplify to ln(32), so you can take out the exponent to have 2ln(3).d/dx(95x)=59049x*2ln(3)*(5)d/dx(95x)=10*59049x*ln(3)
What is the derivative of 2 to the power of 5x?
25x?d/dx(au)=au*ln(a)*d/dx(u)d/dx(25x)=25x*ln(2)*d/dx(5x)-The derivative of 5x is:d/dx(cu)=c*du/dx where c is a constantd/dx(5x)=5*d/dx(x)d/dx(25x)=95x*ln(2)*(5*d/dx(x))-The derivative of x is:d/dx(x)=1x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(25x)=25x*ln(2)*(5*1)d/dx(25x)=25x*ln(2)*(5)-25x can simplify to (25)x, which equals 32x.d/dx(95x)=32x*ln(2)*(5)
What is the slope of the line for the linear equation 5x minus 4y equals 16?
dy/dx = 5/4
Integrate x 5x dx?
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
What is the anti-derivative of 5x-4?
9
How do you find the x- coordinates of the turning points of y equals x3-2x2-5x plus 6?
Differentiate the function with respect to x: d/dx (x3 - 2x2 - 5x + 6) = 3x2 - 4x - 5 Set this derivative = 0 and solve. 3x2 - 4x - 5 = 0 implies that x = -0.7863 or 2.1196 (to 4 dp)
5x3-3x plus 4 -3x3-2x2 plus 6x-1?
Am I right :P?5x^(3)-3x+4-3x^(3)-2x+6x-1Since 5x^(3) and -3x^(3) are like terms, add -3x^(3) to 5x^(3) to get 2x^(3).(d)/(dx) 5x^(3)-3x+4-3x^(3)-2x+6x-1=2x^(3)-3x+4-2x+6x-1Since -3x and -2x are like terms, subtract 2x from -3x to get -5x.(d)/(dx) 5x^(3)-3x+4-3x^(3)-2x+6x-1=2x^(3)-5x+4+6x-1Since -5x and 6x are like terms, subtract 6x from -5x to get x.(d)/(dx) 5x^(3)-3x+4-3x^(3)-2x+6x-1=2x^(3)+x+4-1Subtract 1 from 4 to get 3.(d)/(dx) 5x^(3)-3x+4-3x^(3)-2x+6x-1=2x^(3)+x+3To find the derivative of 2x^(3), multiply the base (x) by the exponent (3), then subtract 1 from the exponent.(d)/(dx) 5x^(3)-3x+4-3x^(3)-2x+6x-1=6x^(2)+(d)/(dx) x+3To find the derivative of x, multiply the base (x) by the exponent (1), then subtract 1 from the exponent (1-1=0). Since the exponent is now 0, x is eliminated from the term.(d)/(dx) 5x^(3)-3x+4-3x^(3)-2x+6x-1=6x^(2)+1+(d)/(dx) 3Since 3 does not contain x, the derivative of 3 is 0.(d)/(dx) 5x^(3)-3x+4-3x^(3)-2x+6x-1=6x^(2)+1+0Add 0 to 1 to get 1.(d)/(dx) 5x^(3)-3x+4-3x^(3)-2x+6x-1=6x^(2)+1The derivative of 5x^(3)-3x+4-3x^(3)-2x+6x-1 is 6x^(2)+1.6x^(2)+1
Integration of x ln 5x dx?
int x ln5x dx by parts u = ln5x du = 1/5x or 5x^-1 dv = x v = 1/2x^2 uv - int v du ln5x 1/2x^2 - int 1/2x^2 5x^-1 1/2ln5x*x^2 - 1/6x^3 5x + C
Integration of ln 5x dx?
5/5x + c where c is the constant of intergration just differentiate the 5x to get 5 and times that by 1/5x then add c The answer above is wrong. This is simply because you cannot intergrate lnx to get 1/x. To intergrate I would recommend using intergration by parts. u=ln5x du/dx=1/x dv/dx=1 v=x uv-(intergal of)v.du/dx =xln5x-intergral of x/x intergral of x/x = x =xln5x-x+c = x(ln5x-1)+c
How do you find the definition of derivative of f of x equals 3x plus 2?
The definition of the derivative, at a point X = x is the limit, as dx tends to 0, of [f(x+dx)-f(x)]/dx. In this case, therefore, it is lim[3*(x+dx)+2 - (3*x+2)]/dx = lim[3x + 3*dx +2 - 3x - 2]/dx = lim[3*dx/dx] = lim[3] = 3.
