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Use the power/chain rules:
d/dx (2 cos2 x) = 2 d/dx (cos x)2 = (4 cos x)*d/dx(cos x) = -4 cos x sin x = -2 sin 2x
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What is the second derivative of x to the fourth power?
d/dx(X^4) = 4X^3 ( first derivative ) d/dx(4X^3) = 12X^2 ( second derivative )
What is the derivative of x-cosx?
Take the derivative term by term. d/dx(X - cosX) = sin(X) ======
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.
What is the derivative with respect to x of the integral of a function of x with respect to x?
d/dx ∫ f(x) dx = f(x)
What is the derivative of ln 1 plus x?
d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)
What is the derivative of ddx(a2x)?
Assuming you mean what is the value of the derivative d/dx(a²x), then: d/dx(a²x) = a² The derivative (with respect to x) of d/dx(a²x) = d/dx(d/dx(a²x)) = d/dx(a²) = 0.
What is the Second derivative of 3.9625.lnx?
3.9625lnx?The first derivative is:d/dx(cu)=c*du/dx where c is a constant.d/dx(3.9625lnx)=3.9625*d/dx(lnx)-The derivative of lnx is:d/dx(lnu)=(1/u)*d/dx(u)d/dx(lnx)=(1/x)*d/dx(x)d/dx(3.9625lnx)=3.9625*[(1/x)*d/dx(x)]-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(3.9625lnx)=3.9625*[(1/x)*1]d/dx(3.9625lnx)=3.9625*(1/x)d/dx(3.9625lnx)=3.9625/xThe second derivative of 3.9625lnx is the derivative of 3.9625/x=3.9625*x-1:d/dx(cu)=c*du/dx where c is a constant.d/dx(3.9625*x-1)=3.9625*d/dx(x-1)-The derivative of x-1 is:d/dx(xn)=nxn-1d/dx(x-1)=-1*x-1-1d/dx(x-1)=-1*x-2d/dx(x-1)=-1/x2d/dx(3.9625*x-1)=3.9625*(-1/x2)d/dx(3.9625*x-1)=-3.9625/x2
Derivative of x?
d/dx x = 1
What is the second derivative of x ln x?
f(x)=xln(x) this function is treated as u*v u=x v=ln(x) The derivative of a product is f'(x)=u*v'+v*u' plugging the values back in you get: f'(x)=(x*dlnx/x)+(ln*dx/dx) The derivative of lnx=1/x x=u dlnu/dx=(1/u)*(du/dx) dx/dx=1 x=u dun/dx=nun-1 dx1/dx=1*x1-1 = x0=1 f'(x)=x*(1/x)+lnx*1 f'(x)=1+lnx Now for the second derivative f''(x)=d1/dx+dlnx/dx the derivative of a constant, such as 1, is 0 and knowing that the derivative of lnx=1/x you get f''(x)=(1/x)
What is the derivative of pi raised to x?
y=x^pid/dx=pi*(x^pi-1)This is true because of power rule.d/dx (x^a)=a(x^(a-1))
What is the derivative of x-y?
The partial derivative in relation to x: dz/dx=-y The partial derivative in relation to y: dz/dy= x If its a equation where a constant 'c' is set equal to the equation c = x - y, the derivative is 0 = 1 - dy/dx, so dy/dx = 1
What is the second derivative of x to the fourth power?
d/dx(X^4) = 4X^3 ( first derivative ) d/dx(4X^3) = 12X^2 ( second derivative )
What is derivative of 1 - x?
d/dx (1-x)=d/dx (1)-d/dx(x)=0-1=-1
What is the derivative of x-cosx?
Take the derivative term by term. d/dx(X - cosX) = sin(X) ======
What is the derivative of 2Y?
It depends: The derivative with respect to Y is: dX/dY 2Y = 2 The derivative with respect to X is: dY/dX 2Y = 0
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.
What is the derivative with respect to x of the integral of a function of x with respect to x?
d/dx ∫ f(x) dx = f(x)
