Find the derivative y 5x4tan3x Find the derivative y In4

Find the derivative: y = 5x4tan3x Find the derivative: y = In(4 - x/4 + x) Find the derivative: y = (4x3 - 3)4. Consider f(x) = x/x + 5. Find the first, second, and third derivatives for f(x). Find dy/dx implicitly: x y2 + y3 x2 = 6

Solution

8 ) xy^2 +y^2 x^2 =0 y^2 + 2x y y\' + 3y^2 y\' x^2 + 2x y^3 =0 y\' ( 2x y + 3y^2 x^2 ) + y^2 + 2x y^3 =0 y\' ( 2x y + 3y^2 x^2 ) = -(y^2 + 2x y^3 ) y\' = dy/dx = -(y^2 + 2x y^3 ) / ( 2x y + 3y^2 x^2 ) 7) f(x) = x / (x+5) = x (x+5)^(-1) f \'(x) = (x+5)^(-1) - x (x+5)^(-1) f \' \' (x) = - (x+5)^(-2) - (x+5)^(-1) + x (x+5)^(-2) f \' \' \' (x) = 2 (x+5)^(-3) + (x+5)^(-2) + (x+5)^(-2) - 2x (x+5)^(-3) 6 -b ) y= (4x^3 -3)^(4) y\' = 4(4x^3 -3)^(4-1) . (4x^3 -3)\' y\' = 4 (4x^3 -3)^(3) . (12x^2) y\' = 28 x^2 (4x^3 -3)^(3) 6-a) y= ln( (4-x)/(4+x) ) u= (4-x)/(4+x) (ln u) \' = u\'/u u= (4-x)/(4+x) ==> u\' = ( -1 (x+4) - 1(4-x)) / (x+4)^2 u\' = ( -x-4 -4+x) / (x+4)^2 u\'= -8 / (x+4)^2 y\' = (ln u) \' = u\'/u = ( -8 / (x+4)^2 ) / ((4-x)/(4+x) )