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

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 y=f(x) is a differentiable function x such that invrse function x=f^(-1) y exists, then prove that x is a differentiable function of y and (dx)/(dy)=(1)/((dy)/(dx)) where (dy)/(dx) ne0 Hence, find (d)/(dx)(tan^(-1)x) .

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

If f(x) and g(x) are two differentiable functions, show that f(x)g(x) is also differentiable such that (d)/(dx)[f(x)g(x)]=f(x)(d)/(dx){g(x)}+g(x)(d)/(dx){f(x)}

If f(x) and g(x) a re differentiate functions, then show that f(x)+-g(x) are also differentiable such that (d)/(dx){f(x)+-g(x)}=(d)/(dx){f(x)}+-(d)/(dx){g(x)}

If x=f(t), y=g(t) are differentiable functions of parameter 't' then prove that y is a differentiable function of 'x' and (dy)/(dx)=((dy)/(dt))/((dx)/(dt)),(dx)/(dt) ne 0 Hence find (dy)/(dx) if x=a cos t, y= a sin t.

If f(x) and g(x) are differentiable function then show that f(x)+-g(x) are also differentiable such that d(f(x)+-g(x))/(dx)=d(f(x))/(dx)+-d(g(x))/(dx)

If x=f(t) and y=g(t) are differentiable functions of t then (d^(2)y)/(dx^(2)) is

If x and y are differentiable functions of t, then (dy)/(dx)=(dy//dt)/(dx//dt)," if "(dx)/(dt)ne0 .