#### To determine

**To find:** The derivative of y=(x+1x)5.

#### Answer

The derivative of y=(x+1x)5 is y′=5(x2+1)4(x2−1)x6_.

#### Explanation

**Given:**

The function is y=(x+1x)5.

**Result used:**

**The Chain Rule:**

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

F′(x)=g′(h(x))⋅h′(x) (1)

**Derivative Rules:**

(1) Power Rule: ddx(xn)=nxn−1.

(2) Sum Rule: ddx[f(x)+g(x)]=ddx[f(x)]+ddx[g(x)].

**Calculation:**

Obtain the derivative of *y.*

y′=ddx(y)=ddx((x+1x)5)

Let h(x)=x+1x and g(u)=(u)5 where u=h(x).

Apply the chain rule as shown in equation (1),

y′=g′(h(x))⋅h′(x) (2)

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

g′(h(x))=g′(u) [Qu=h(x)]=ddu(g(u))=ddu(u)5

Apply the power rule (1) and simplify further.

g′(h(x))=(5u5−1)=5u4=5(x+1x)4 [Qu=x+1x]

Thus, the derivative is g′(h(x))=5(x+1x)4 (3)

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

h′(x)=ddx(x+1x)=ddx(x+x−1)

Apply the sum rule (2) and the power rule (1),

h′(x)=ddx(x)+ddx(x−1)=(1x1−1)+(−1x−1−1)=1−x−2=1−1x2

Thus, the derivative is h′(x)=1−1x2 (4)

Substitute equations (3) and (4) in equation (2),

y′=5(x+1x)4⋅(1−1x2)=5(x2+1x)4(x2−1x2)=5(x2+1)4(x2−1)x4⋅x2=5(x2+1)4(x2−1)x6

Therefore, the derivative of y=(x+1x)5 is 5(x2+1)4(x2−1)x6_.