derivatives dy/dx of y=(sinx)^(2sinx)?

Question

don't understand how to derive this.

Answer 1

For differentiating the function y = (sinx)^(2sinx) first take Ln on both sides, applying the rules applicable to Ln, getting:

Ln (y) = 2sinx*Ln(sinx).

Now differentiate both sides w.r.t. x, using the product formula on R.H.S.

1/y*dy/dx = 2sinx*[(1/sinx)*cosx] + Ln(sinx)*2cosx

= 2cosx + Ln(sinx)*2cosx = 2cosx[1 + Ln(sinx)]

or dy/dx = y×2cosx[1 + Ln(sinx)]

or dy/dx = [(sinx)^(2sinx)]×2cosx[*1 + Ln(sinx)].

Answer 2

Take ln
y= sinx)^2sinx)
lny = 2 sin x * ln sin x
y´/y= 2[cosx*ln sin x +sin x*1/sinx *cos x)
so y´= (sin x)^2 sin x * cos x( ln sin x +1)*2

Answer 3

(x²/sixteen) - (y²/9) = a million (x/4)² - (y/3)² = a million [(x/4) - (y/3)]*[(x/4) + (y/3)] = a million [(3x - 4y)/12]*[(3x+ 4y)/12] = a million [(3x - 4y)]*[(3x+ 4y)] = 12 9x² - 16y² = 12 6y² = 9x² - 12 y² = (3/2)x² - 12 <-----eq. a million differentiate the two facets of equation a million with comprehend to variable x 2*y*(dy/dx) = (3/2)*2*x*(dx/dx) 2y(dy/dx) = (3)*(x)*(a million) (dy/dx) = 3x/2y (dy/dx) = (3/2)(x/y) <----- equation 2 remedy for y from equation a million y² = (3/2)x² - 12 y = ?((3/2)x² - 12) plug interior the fee of y in terms of x to equation 2 (dy/dx) = (3/2)(x/y) (dy/dx) = (3/2)(x/(?((3/2)x² - 12)))