Tangent line and derivative?

Question

.1) Find dy/dx when y=x^(4lnx) 

2. find the slope of the tangent line equation of the curve x^(2) ln (y)+xy^(3)=5 at poin (5,1) on the curve by using implicit differentiation

Answer 1

2)

x² ln(y) + xy³ = 5

Apply derivative both sides with respect to x.

(d/dx) [ x² ln(y) + xy³ ] = (d/dx) 5

(d/dx) [ x² ln(y) ] + (d/dx) [ xy³ ] = 0

Apply formula : (d/dx)(UV) = Udv + VdU

x² (d/dx)[ln(y)] + ln(y)(d/dx)[x²]+ x(d/dx)[y³] +y³(d/dx)[x] = 0

x²(1/y)y'+ ln(y)2x+ x(3y²)y' +y³ = 0

(x²/y)y'+ (2x)ln(y)+ 3xy²y' + y³ = 0

(x²/y)y'+ 3xy²y' + (2x)ln(y) + y³ = 0

[ (x²/y)+ 3xy² ]y' = - (2x)ln(y) - y³

y' = [ - (2x)ln(y) - y³ ] / [ (x²/y)+ 3xy² ]

Slope of tangent line is dy/dx = y' = [ - (2x)ln(y) - y³ ] / [ (x²/y)+ 3xy² ]  which is passes through the point (5,1)

y' = [ - (2*5)ln(1) - 1³ ] / [ (5²/1)+ 3(5)(1²) ]

y' = [ - (2*5)(0) - 1 ] / [ (25)+ 15 ]

y' = - 1/ 40

The slope of the given tangent line is -1/40.