#### To determine

**To find:** The derivative of f(x)g(x)=(f(x))(g(x))−1.

#### Answer

The derivative of f(x)g(x)=(f(x))(g(x))−1 is ddx[f(x)g(x)]=g(x)[f′(x)]−f(x)[g′(x)][g(x)]2_.

#### Explanation

**Result used:** *Chain Rule*

If *g* is differentiable at *x* and *f* is differentiable at g(x), then the composite function F=f∘g defined by F(x)=f(g(x)) is differentiable at *x* and F′ is given by the product

F′(x)=f′(h(x))⋅h′(x) (1)

**Derivative Rule:**

(1) *Power Rule*: ddx(xn)=nxn−1.

(2) *Product Rule*: ddx[f(x)g(x)]=f(x)ddx[g(x)]+g(x)ddx[f(x)]

**Calculation:**

Obtain the derivative of f(x)g(x)=f(x)(g(x))−1.

ddx(f(x)g(x))=ddx[(f(x))(g(x))−1]

Apply the product rule (2),

ddx(f(x)g(x))=f(x)ddx[(g(x))−1]+(g(x))−1ddx(f(x))

=f(x)ddx[(g(x))−1]+(g(x))−1(f′(x)) (2)

Obtain the derivative ddx[(g(x))−1] using the chain rule as shown in equation (1).

Let h(x)=g(x) and f(u)=u−1 where u=h(x).

ddx[(g(x))−1]=f′(h(x))⋅h′(x) (3)

The derivative f′(h(x)) is computed as follows,

f′(h(x))=f′(u)=ddu(f(u))=ddu(u−1)

Apply derivative rule (1),

f′(h(x))=(−1)u−1−1=−u−2=−1u2

Substitute u=g(x) in the above equation,

f′(h(x))=−1(g(x))2

Thus, the derivative is f′(h(x))=−1(g(x))2.

The derivative of h(x) is computed as follows,

h′(x)=ddx(g(x))=g′(x)

Thus, the derivative of h(x) is h′(x)=g′(x).

Substitute −1(g(x))2 for f′(h(x)) and g′(x) for h′(x) in equation (3),

ddx[(g(x))−1]=−1(g(x))2(g′(x))=−g′(x)(g(x))2

Substitute ddx[(g(x))−1]=−g′(x)(g(x))2 in equation (2),

ddx(f(x)g(x))=f(x)(−g′(x)(g(x))2)+(g(x))−1(f′(x))=−f(x)(g′(x))(g(x))2+(g(x))−1(f′(x))=(g(x))−1(g(x))2(f′(x))−f(x)(g′(x))(g(x))2=g(x)(f′(x))−f(x)(g′(x))(g(x))2

Therefore, the derivative of f(x)g(x)=(f(x))(g(x))−1 is ddx[f(x)g(x)]=g(x)[f′(x)]−f(x)[g′(x)][g(x)]2_.