#### To determine

**To find:** The derivative of the function U(y)=(y4+1y2+1)5.

#### Answer

The derivative of U(y)=(y4+1y2+1)5 is 10y(y4+1)4(y4+2y2−1)(y2+1)6_.

#### Explanation

**Given:**

The function is U(y)=(y4+1y2+1)5.

**Result used:**

**The Chain Rule:**

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

F′(y)=g′(h(y))⋅h′(y) (1)

**Derivative Rule:**

(1) Power Rule: ddx(xn)=nxn−1.

(2) Quotient Rule: ddy[f(y)g(y)]=g(y)ddy[f(y)]−f(y)ddy[g(y)][g(y)]2

**Calculation:**

Obtain the derivative of U(y).

U′(y)=ddy(U(y))=ddy((y4+1y2+1)5)

Let h(y)=y4+1y2+1 and g(v)=v5 where v=h(y)

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

U′(y)=g′(h(y))⋅h′(y) (2)

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

g′(h(u))=g′(v)=ddv(g(v))=ddvv5

Apply the power rule (2) then substitute v=y4+1y2+1,

g′(h(y))=5v5−1=5v4=5(y4+1y2+1)4

Thus, the derivative is g′(h(y))=5(y4+1y2+1)4.

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

h′(y)=ddy(y4+1y2+1)

Apply the Quotient Rule as shown in equation (3),

h′(y)=(y2+1)ddy[y4+1]−(y4+1)ddy[y2+1](y2+1)2=(y2+1)[ddy(y4)+ddy(1)]−(y4+1)[ddy(y2)+ddy(1)](y2+1)2=(y2+1)[(4y4−1)+(0)]−(y4+1)[(2y2−1)+(0)](y2+1)2=(y2+1)[(4y3)]−(y4+1)[(2y)](y2+1)2

Simplify further and obtain the derivative.

h′(y)=4y3(y2+1)−2y(y4+1)(y2+1)2=(4y5+4y3)−(2y5+2y)(y2+1)2=4y5+4y3−2y5−2y(y2+1)2=2y5+4y3−2y(y2+1)2

Thus, the derivative of h(y) is h′(y)=2y5+4y3−2y(y2+1)2.

Substitute 5(y4+1y2+1)4 for g′(h(y)) and 2y5+4y3−2y(y2+1)2 for h′(y) in equation (2),

g′(h(y))⋅h′(y)=5(y4+1y2+1)4⋅2y5+4y3−2y(y2+1)2=5(y4+1)4(y2+1)4⋅2y(y4+2y2−1)(y2+1)2=10y(y4+1)4(y4+2y2−1)(y2+1)4+2=10y(y4+1)4(y4+2y2−1)(y2+1)6

Therefore, The derivative of U(y)=(y4+1y2+1)5 is U′(y)=10y(y4+1)4(y4+2y2−1)(y2+1)6_.