# differentiation-3

3) _{ } Why?

Proof:

Let y = f_{1}(x) + f_{2}(x), Where f_{1}(x) and f_{2}(x) are differentiable functions of x.

Then

_{ }

Thus _{ } .

4) _{ } Why?

Proof:

Let y = f_{1}(x).f_{2}(x) where both f_{1}(x) and f_{2}(x) are differentiable function of x.

_{ }

_{ }

_{ }

_{ }

5) _{ } Why?

Proof:

Let_{ } , where f_{1}(x) and f_{2}(x) are differentiable functions at all points in its domain and f_{2}(x) ¹ 0.

_{ }

_{ }

6) _{ } Why?

Proof:

Let y = f (t), t = g(x)

Then y = f (g(x)) is a function of x.

Dy = f (t+Dt) - f (t)

And Dt = g (x+Dx) - g(x)

_{ }

Assuming that for sufficient small values of Dx, Dt ¹ 0 this will necessarily be the case if _{ } then since g is differentiable it is continuous and hence when Dx®0, x+Dx®x any g(x+Dx) ® g(x).

Therefore t+Dt®t and Dt®0, since Dt ¹ 0 and Dx®0 we may write,

_{ } .

Clearly f and g are both continuous functions because they are differentiable thus Dt®0 when Dx®0 and Dy®0 when Dt®0.

_{ }

Hence_{ }

**Standard Formulae of Differentiation: **

1) _{ }

Proof:

Let _{ }

_{ }

_{ }

Thus_{ }

2) _{ }

Proof:

_{ }

Thus_{ }

3) _{ }

Proof:

_{ }

_{ }

So, _{ }

4) _{ }

Proof:

_{ }