Open Question: Y^2 = 3*(x^3)*sinx . Could someone please help me to find dy/dx?
8:02 Publicado por Flechado16
Y^2 = 3*(x^3)*sinx
Y= v(3*(x³sinx))
= (v3) (x^(3/2)* (sin^½x)
Y= v(3*(x³sinx))
= (v3) (x^(3/2)* (sin^½x)
U=x^(3/2); V = sin^(½)x
Dy/dx= v3(U.V' + VU')
= v3[x^3/2*½ (sin^-½x)cosx + sin^(½)x*(3/2x^½)]
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