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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
What is the anti-derivative of 5x-4?
9
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
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
What is the derivative of 5 divided by x2?
2.5
What is the derivative of 5xy?
To find the derivative of the expression (5xy) with respect to (x), we can use the product rule. The derivative is given by ( \frac{d}{dx}(5xy) = 5 \left( x \frac{dy}{dx} + y \right) ). Thus, the derivative of (5xy) is (5y + 5x \frac{dy}{dx}).
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
