If u and v are two functions of x then prove that: `intuv dx =uint vdx - int [(du)/(dx)int vdx] dx`

If u and v are two functions of x then prove that : int uv dx = u int v dx - int [ (du)/(dx) int v dx ] dx Hence, evaluate int x e^(x) dx

If u and v are integrable function of x, then, show that intu.v.dx=uintvdx-int[(du)/(dx)intvdx]dx . Hence evaluate intlogxdx .

If u and v are two differentiable functions of x and int uv dx=phi(x)-int[(du)/(dx)intv dx]dx , then the value of phi(x) is-

U=x^4 , Find (du)/dx

If u, v and w are functions of x, then show that d/dx(u cdot v cdot w) = du/dx (v cdotw + u cdot dv/dx cdot w +u cdot 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. Using product rule

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.