Home
Class 12
MATHS
If y =f(u) is differentiable function of...

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)`.

Text Solution

Verified by Experts

Let `deltau` and `deltay` be the increments in u and y respectively, corresponding to the increment `deltax` in x.
Now y is a differentiable function of u and u is a differentiable function of x.
`therefore (dy)/(du)=underset(deltaurarr0)lim(deltay)/(deltau)and (du)/(dx)=underset(deltaxrarr0)lim(deltau)/(deltax)" ... (1)"`
Also `underset(deltaxrarr0)limdeltau=underset(deltaurarr0)lim((deltau)/(deltax).deltax)`
`=(underset(deltaxrarr0)lim(deltau)/(deltax))(underset(deltaxrarr0)limdeltax)=(du)/(dx)xx0=0`
This means that as `deltaxrarr0, deltaurarr0" ... (2)"`
`(deltay)/(deltax)=(deltay)/(deltau)xx(deltau)/(deltax)" ..."[deltaune0]`
Taking limits as `deltax rarr0`, we get,
`underset(deltaxrarr0)lim(deltay)/(deltax)=underset(deltaxrarr0)lim((deltay)/(deltau)xx(deltau)/(deltax))`
`=underset(deltaxrarr0)lim(deltay)/(deltau)xxunderset(deltaxrarr0)lim(deltau)/(deltax)`
`=underset(deltaurarr0)lim(deltay)/(deltau)xxunderset(deltaxrarr0)lim(deltau)/(deltax)" ... [By (2)]"`
Now both the limits on RHS exist ... [By (1)]
`therefore underset(deltaxrarr0)lim(deltay)/(deltax)` exists and is equal to `(dy)/(dx)`.
`therefore (dy)/(dx)=(dy)/(du)xx(du)/(dx)" ... [By (1)]"`
Promotional Banner

Topper's Solved these Questions

  • DIFFERENTIATION

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise Ex. 1 Solved Examples|4 Videos

  • DIFFERENTIATION

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise Ex. 2 Solved Examples|3 Videos

  • DIFFERENTIAL EQUATIONS

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise M.C.Q|12 Videos

  • INTEGRATION

    NAVNEET PUBLICATION - MAHARASHTRA BOARD|Exercise Multiple Choice Questions|15 Videos

Similar Questions

Explore conceptually related problems

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) .

Differential equations of the type (dy)/(dx)=f(x)

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

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