# If g is continuous on a and f(x)=((x^2)-(a^2))g(x). Determine f'(a).

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### 2 Answers

You need to use the definition of derivative at a poin `x = a` , such that:

`f'(a) = lim_(x->a)(f(x) - f(a))/(x - a)`

`f'(a) = lim_(x->a)((x^2 - a^2)*g(x) - (a^2 - a^2)*g(a))/(x - a)`

Reducing duplicate terms yields:

`f'(a) = lim_(x->a)(x^2 - a^2)*(g(x))/(x - a)`

Converting the difference of squares into a product yields:

`f'(a) = lim_(x->a)(x - a)(x + a)*(g(x))/(x - a)`

Reducing duplicate terms yields:

`f'(a) = lim_(x->a)(x + a)*(g(x))`

`f'(a) = 2a*g(a)`

**Hence, evaluating the derivative of the function f(x) at x = a, yields **`f'(a) = 2a*g(a).`

First we not that f(x) is continue too, as product of two functons `(x^2-a^2)` and `g(x) ` contiunes.

Now if we suppose that `I ` is the neighbour of `a` where `g(x)` is contiunos, since `(x^2-a^2)` is continue on all over `R` then in `I` `f(x)` is contiunue too. So if exists a limit `b< oo` :

`lim_(x->a) (x^2-a^2)g(x)=b` `=f(a)`

Derivative is:

`lim_(x->a) (f(x)-f(a))/(x-a)=lim_(x->a) (x^2-a^2)(g(x)-g(a))/(x-a)=`

`=lim_( x->a) (x+a)(x-a)(g(x)-g(a))/(x-a)=``=lim_(x->a) (x+a)(g(x)-b)=` `2a lim_(x->a) (g(x)-b)=0`

Note that it is not clear that: `lim_(x->a)(x^2-a^2)(g(x)-g(a))/(x-a)=0`

for `g(x)` is continue on `I` according hypothesis, but not limited.

We have added the hypotesis `lim_(x->a) (x^2-a^2) g(x)= b`

so that `g(x) ` is dominated by `(x^2-a^2) ` in `I` .and so make us sure exists derivative of `f(x)` in `a` by passing the fact we dunno what realy `g(x) ` behaviour is. Actually we know only `f(x)` is contiune in `a` as product of two contiunue functions, nothing short of.

Indeed let ya see `g(x)= (log^2(x/a))/(x-a)^2`

then: `lim_(x->a) g(x)=lim_(x->a) (log(x/a) a/x)/(x-a)=`

`=lim_(x->a) a/x xx lim_(x->a) a/x = 1`

We have applied De L'Hospital's rule a twice.

Thus exists `lim_(x->a) g(x)`

On the other side:

`lim_(x->a) (x^2-a^2)(g(x)-g(a))/(x-a)^3=lim_(x->a) ((x+a)(log^2(a/x)-1))/(x-a)^2 `

does'nt exists and neither `f'(x)`

So the condition of merely limit isn't enough to assure existence of the limt of product of two funtion , even contiunes,and rtheir relative derivative existence.