Logarithmic Differentiation Help NeededA LOT OF PROCESS WORK IS SHOWN

[johnq2k7]

Use logarithmic differentiation to find:

a.) d/dx of [(sin^-1(x^2)*sinh^-1(x^2))/(sin^4(x^2))]

b.) d^2/dx^2 (sech^-1(e^(2*x)))


work shown for a:

let y= [(sin^-1(x^2)*sinh^-1(x^2))/(sin^4(x^2))]

taking the natural logarithm of both sides:

ln y= ln [(sin^-1(x^2)*sinh^-1(x^2))/(sin^4(x^2))]

ln y= ln (sin^-1(x^2)*sinh^-1(x^2)) - ln (sin^4(x^2))

ln y = ln (sin^-1(x^2)) +ln (sinh^-1(x^2)) - 4*ln(sin^4(x^2))

differentiating both sides:

1/y* y'= (1/sin^-1(x^2))(1/sqrt(1-(x^2)^2)(2x) + (1/sinh^-1(x^2))(2x)(1/sqrt(1+(x^2)^2)) - (4)(1/sin(x^2))(2x*cos(x^2))

substituting y i got:

y'= (sin^4(x^2)*2x)/(sqrt(1-x^4)) + 2xsin^4(x^2)/(sqrt(1+x^4)) - 8x*cos(x^2)sin^3(x^2)

is this correct... i think I may have made a few errors

work shown for b:

let y= sech^-1(e^(2x))

taking natural logarithm of both sides:

ln y= ln (sech^-1(e^(2x))

differentiating both sides:

y'= 2/(sqrt(1-e^(4x))

to find d^2/dx^2 i set dy/dx as y again:

therefore let y= 2/(sqrt(1-e^(4x))

finding the natural logarithm of both sides of dy/dx

i get ln y= ln 2- (1/2)ln (1-e^(4x))

differentiating both sides leads to:

y"= 4e^(4x)/((1-e^(4x))^(3/2))

is this correct.. i believe i may have made a few errors.. please check

 

[badphysicist]

I looked over a and I think that it is pretty good, but I think you made errors when you plugged y back in.