# Product Rule

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**Product Rule**. If `f` and `g` are both differentiable then `(f(x)g(x))'=f(x)g'(x)+f'(x)g(x)`.

Proof.

By definition

`(f(x)g(x))=lim_(h->0)(f(x+h)g(x+h)-f(x)g(x))/h=`

`=lim_(h->0)(f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)-f(x)g(x))/h=`

`=lim_(h->0)(f(x+h)g(x+h)-f(x+h)g(x))/h+lim_(h->0)(f(x+h)g(x)-f(x)g(x))/h=`

`=lim_(h->0)(f(x+h)(g(x+h)-g(x))/h)+lim_(h->0)(g(x)(f(x+h)-f(x))/h)=`

`lim_(h->0)f(x+h)lim_(h->0)(g(x+h)-g(x))/h+lim_(h->0)g(x)lim_(h->0)(f(x+h)-f(x))/h=` by product law of limits (can be used because limits exist)

`=f(x+0)g'(x)+g(x)f'(x)=f(x)g'(x)+f'(x)g(x)` .

**Example 1**. Find derivative of `f(x)=xe^x` .

`f(x)'=(xe^x)'=x(e^x)'+(x')e^x=xe^x+1*e^x=e^x(x+1)` .

**Example2**. Find derivative of `f(t)=t^2sqrt(t)`

`f'(t)=(t^2sqrt(t))'=t^2(sqrt(t))'+(t^2)'sqrt(t)=t^2 1/(2sqrt(t))+2tsqrt(t)=5/2tsqrt(t)` .

Note, however that Example 2 can be solved without product rule. Since `t^2sqrt(t)=t^2t^(1/2)=t^(5/2)` then `f'(t)=5/2 t^(5/2-1)=5/2 t^(3/2)`.

Example 2 showed that first you need to look at the function and make sure that product rule is the only way to find derivative (like in Example 1).

Product rule can be extended on case of more than two functions.

Let's see how this can be done for three functions:

`(uvw)'=(uv)'w+uvw'=(u'v+uv')w+uvw'=u'vw+uv'w+uvw'`.

Similarly can be proven case for four functions:

`(uvwz)'=u'vwz+uv'wz+uvw'z+uvwz'`.

Do you see pattern? In general if we have `n` functions then derivative of product is sum of `n` products and in each product only one function is differentiated.

**Example 3**. Differentiate `y=x^2sin(x)e^x`.

`y'=(x^2sin(x)e^x)'=(x^2)'sin(x)e^x+x^2(sin(x))'e^x+x^2sin(x)(e^x)'=`

`=2xsin(x)e^x+x^2cos(x)e^x+x^2sin(x)e^x=xe^x(2sin(x)+xcos(x)+xsin(x))`.

**Example 4**. Find `(x^2sin(x)ln(x)e^x)'`.

`(x^2sin(x)ln(x)e^x)'=`

`=(x^2)'sin(x)ln(x)e^x+x^2(sin(x))'ln(x)e^x+x^2sin(x)(ln(x))'e^x+x^2sin(x)ln(x)(e^x)'=`

`=2xsin(x)ln(x)x^2cos(x)ln(x)e^x+xsin(x)e^x+x^2sin(x)ln(x)e^x`.