Prove that `(d)/(dx)uv=u(dv)/(dx)+v(du)/(dx)`.

If u,v and w are functions of x ,then show that (d)/(dx)(u*v*w)=(du)/(dx)v*w+u*(dv)/(dx)*w+u*v(dw)/(dx) in two ways - first by repeated application of product rule,second by logarithmic differentiation.

If u, v and w are functions of x, then show that (d)/(dx)(u.v.w) = (du)/(dx) v.w+u. (dv)/(dx).w+u.v(du)/(dx) in two ways-first by repeated application of product rule, second by logarithmic differentiation.

If u, v and w are functions of x, then show that (d)/(dx)(u.v.w) = (du)/(dx) v.w+u. (dv)/(dx).w+u.v(dw)/(dx) in two ways-first by repeated application of product rule, second by logarithmic differentiation.

If u, v and w are functions of x, then show that (d)/(dx)(u.v.w) = (du)/(dx) v.w+u. (dv)/(dx).w+u.v(dw)/(dx) in two ways-first by repeated application of product rule, second by logarithmic differentiation.

If u, v and w are functions of x, then show that (d)/(dx)(u.v.w) = (du)/(dx) v.w+u. (dv)/(dx).w+u.v(dw)/(dx) in two ways-first by repeated application of product rule, second by logarithmic differentiation.

If y =f(u) is differentiable function of u, and u=g(x) is a differentiable function of x, then prove that y= f [g(x)] is a differentiable function of x and (dy)/(dx)=(dy)/(du)xx(du)/(dx) .

If y = f(u) is a differentiable functions of u and u = g(x) is a differentiable functions of x such that the composite functions y = f[g(x) ] is a differentiable functions of x then (dy)/(dx) = (dy)/(du) . (du)/(dx) Hence find (dy)/(dx) if y = sin ^(2)x

If u, v and w are functions of x, then show that (d)/(dx) (u.v.w) = (du)/(dx) v.w + u.(dv)/(dx).w + u.v.(dw)/(dx) in two ways- first by repeated application of product rule, second by logarithmic differentiation. Using product rule