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Is Dx Over

Updated: 12/15/2022
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Q: Is Dx Over
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Calculate dy over dx when xy3 squared plus y equals 3x?

Calculate dx when xy3 + y = 3x.


Derivative of 5ex plus 2?

5ex+2?d/dx(u+v)=du/dx+dv/dxd/dx(5ex+2)=d/dx(5ex)+d/dx(2)-The derivative of 5ex is:d/dx(cu)=c*du/dx where c is a constant.d/dx(5ex)=5*d/dx(ex)-The derivative of 2 is 0 because it is a constant.d/dx(5ex+2)=(5*d/dx(ex))+(0)d/dx(5ex+2)=5*d/dx(ex)-The derivative of ex is:d/dx(eu)=eu*d/dx(u)d/dx(ex)=ex*d/dx(x)d/dx(5ex+2)=5*(ex*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(5ex+2)=5*(ex*1)d/dx(5ex+2)=5*(ex)d/dx(5ex+2)=5ex5ex+2?d/dx(cu)=c*du/dx where c is a constant.d/dx(5ex+2)=5*d/dx(ex+2)-The derivative of ex+2 is:d/dx(eu)=eu*d/dx(u)d/dx(ex+2)=ex+2*d/dx(x+2)d/dx(5ex+2)=5*(ex+2*d/dx(x+2))-The derivative of x+2 is:d/dx(u+v)=du/dx+dv/dxd/dx(x+2)=d/dx(x)+d/dx(2)d/dx(5ex+2)=5*[ex+2*(d/dx(x)+d/dx(2))]-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)=1-The derivative of 2 is 0 because it is a constant.d/dx(5ex+2)=5*[ex+2*(1+0)]d/dx(5ex+2)=5*[ex+2*(1)]d/dx(5ex+2)=5*[ex+2]d/dx(5ex+2)=5ex+2


What is the derivative of 1 over 2 x?

d/dx(1/2x) = -1/(2x2)


What is the derivative of lnx raised to 4?

ln(x4)?d/dx(ln(u))=1/u*d/dx(u)d/dx(ln(x4))=[1/x4]*d/dx(x4)-The derivative of x4 is:d/dx(x4)=4x4-1d/dx(x4)=4x3d/dx(ln(x4))=[1/x4]*(4x3)d/dx(ln(x4))=4x3/x4d/dx(ln(x4))=4/x(lnx)4?Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(lnx)4=4(lnx)4-1*d/dx(lnx)d/dx(lnx)4=4(lnx)3*d/dx(lnx)-The derivative of lnx is:d/dx(ln(u))=1/u*d/dx(u)d/dx(lnx)=1/x*d/dx(x)d/dx(lnx)=1/x*(1)d/dx(lnx)=1/xd/dx(lnx)4=4(lnx)3*(1/x)d/dx(lnx)4=4(lnx)3/x


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)

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The product rule states: d/dx uv = vdu/dx + udv/dxIt is not quite clear what your denominator is:"over 1 + (x squared)": d/dx (x/1+x2)d/dx(x/1+x2) = d/dx x(1 + x2)-1 = (1 + x2)-1 d/dx x + x d/dx (1 + x2)-1= (1 + x2)-1 + x -2x (1 + x2)-2= (1 + x2)-2 (1 + x2 - 2x2)= (1 - x2) / (1 + x2)2"over (1 + x) [all] squared": d/dx (x/(1+x)2)d/dx(x/(1+x)2) = d/dx x(1 + x)-2 = (1 + x)-2 d/dx x + x d/dx (1 + x)-2= (1 + x)-2 + x -2 (1 + x)-3= (1 + x)-3 (1 + x - 2x)= (1 - x)/(1 + x)3If you prefer, you can use the quotient rule: d/dx (u/v) = (v du/dx - udv/dx) / v2


What is the derivative of x to the 4th over 4?

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Derivative of 5ex plus 2?

5ex+2?d/dx(u+v)=du/dx+dv/dxd/dx(5ex+2)=d/dx(5ex)+d/dx(2)-The derivative of 5ex is:d/dx(cu)=c*du/dx where c is a constant.d/dx(5ex)=5*d/dx(ex)-The derivative of 2 is 0 because it is a constant.d/dx(5ex+2)=(5*d/dx(ex))+(0)d/dx(5ex+2)=5*d/dx(ex)-The derivative of ex is:d/dx(eu)=eu*d/dx(u)d/dx(ex)=ex*d/dx(x)d/dx(5ex+2)=5*(ex*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(5ex+2)=5*(ex*1)d/dx(5ex+2)=5*(ex)d/dx(5ex+2)=5ex5ex+2?d/dx(cu)=c*du/dx where c is a constant.d/dx(5ex+2)=5*d/dx(ex+2)-The derivative of ex+2 is:d/dx(eu)=eu*d/dx(u)d/dx(ex+2)=ex+2*d/dx(x+2)d/dx(5ex+2)=5*(ex+2*d/dx(x+2))-The derivative of x+2 is:d/dx(u+v)=du/dx+dv/dxd/dx(x+2)=d/dx(x)+d/dx(2)d/dx(5ex+2)=5*[ex+2*(d/dx(x)+d/dx(2))]-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)=1-The derivative of 2 is 0 because it is a constant.d/dx(5ex+2)=5*[ex+2*(1+0)]d/dx(5ex+2)=5*[ex+2*(1)]d/dx(5ex+2)=5*[ex+2]d/dx(5ex+2)=5ex+2


What is the derivative of x5lnx?

x5lnx?d/dx (uv)=u*dv/dx+v*du/dxd/dx (x5lnx)=x5*[d/dx(lnx)]+lnx*[d/dx(x5)]-The derivative of lnx is:d/dx(lnu)=(1/u)*[d/dx(u)]d/dx(lnx)=(1/x)*[d/dx(x)]d/dx(lnx)=(1/x)*[1]d/dx(lnx)=(1/x)-The derivative of x5 is:d/dx (xn)=nxn-1d/dx (x5)=5x5-1d/dx (x5)=5x4d/dx (x5lnx)=x5*[1/x]+lnx*[5x4]d/dx (x5lnx)=[x5/x]+5x4lnxd/dx (x5lnx)=x4+5x4lnx


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