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**Hint:**Here we will use the basic definition of the differential function. Then we will form the condition of a continuous function. We will then show that the function is continuous to prove that every differentiable function is a continuous function.

**Complete step-by-step answer:**Let \[f\] be the differentiable function at \[x = a\].

Then according to the basic definition of the differentiation, Differentiation of a function is equals to

\[f'\left( c \right) = \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\]……………………….\[\left( 1 \right)\]

We know that the for a function to be continuous at a point it must satisfy the equation

\[\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\]

We can write the above equation as

\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = 0\]……………………\[\left( 2 \right)\]

So, for a function to be continuous it must satisfy the equation \[\left( 2 \right)\].

Now we will find the value of \[\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right)\] for the given differentiable function.

Therefore we can write \[\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right)\] as,

\[\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = \mathop {\lim }\limits_{x \to a} \left( {\dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\left( {x - a} \right)} \right)\]

\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = \mathop {\lim }\limits_{x \to a} \left( {\dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}} \right) \times \mathop {\lim }\limits_{x \to a} \left( {x - a} \right)\]

By using the equation \[\left( 1 \right)\] in the above equation, we get

\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times \mathop {\lim }\limits_{x \to a} \left( {x - a} \right)\]

By putting the limit on the RHS of the equation, we get

\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times \left( {a - a} \right)\]

\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times 0\]

\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = 0\]

Hence as per the condition of the equation \[\left( 2 \right)\] we can say that the given function \[f\] is a continuous function.

**Hence, every differentiable function is continuous.**

**Note:**Here we have to note that continuous function is the function whose value does not change or value remains constant. When the function is continuous at a point then the left hand limit of the function and the right hand limit of the function is equal to the value of the function at that point.

\[\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)\]

Also, a differentiable function is always continuous but the converse is not true which means a function may be continuous but not always differentiable. A differentiable function may be defined as is a function whose derivative exists at every point in its range of domain.

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