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## Homework Statement

I just need to find the derivatives of the following:

[a] y=ln(x^4 sin^2 x)

**f(x) = x^2 lnx, also find f'(1)**

[c] x^y = y^x

See above

[a] y' = [(4x^3)(sin^2x) + (x^4)(2sinxcosx)]/(x^4sin^2x) = (4sinx + 2xcosx)/(xsinx)

[c] x^y = y^x

## Homework Equations

See above

## The Attempt at a Solution

[a] y' = [(4x^3)(sin^2x) + (x^4)(2sinxcosx)]/(x^4sin^2x) = (4sinx + 2xcosx)/(xsinx)

**f(x)' = (2x)(lnx) + (x^2)(1/x) = 2xlnx + x**

f'(1) = 2(1)ln(1) + 1 = 2(0) + 1 = 1

[c] ln(x^y) = ln(y^x)

ylnx = xlny

d/dx ( ylnx) = d/dx (xlny)

(1)(y')(lnx) + (y)(1/x) = (1)(lny) + (x)(1/y)(y')

lnxy' + y/x = lny + xy'/y

lnxy' - xy'/y = lny = y/x

y'(lnx - x/y) = lny - y/x

y' = (lny - y/x)/(lnx - x/y)f'(1) = 2(1)ln(1) + 1 = 2(0) + 1 = 1

[c] ln(x^y) = ln(y^x)

ylnx = xlny

d/dx ( ylnx) = d/dx (xlny)

(1)(y')(lnx) + (y)(1/x) = (1)(lny) + (x)(1/y)(y')

lnxy' + y/x = lny + xy'/y

lnxy' - xy'/y = lny = y/x

y'(lnx - x/y) = lny - y/x

y' = (lny - y/x)/(lnx - x/y)