#### To determine

**To find:** The differential of the function.

#### Answer

The differential of the function is
dy=−4x(x2−3)3dx.

#### Explanation

**Given:**

The function is
y=(x2−3)−2.

**Derivative rules:**

(1) Chain rule: If
y=f(u) and
u=g(x) are both differentiable function, then

dydx=dydu⋅dudx.

(2) Product rule:
ddx(fg)=fddx(g)+gddx(f)

(3). Power Rule:
ddx(xn)=nxn−1

(4) Difference Rule:
ddx[f(x)−g(x)]=ddx[f(x)]−ddx[g(x)]

**Result used:**

If
y=f(x), then the differential is
dy=f′(x)dx.

**Calculation:**

The differential of the function is computed as follows,

Consider
y=f(x).

Differentiate
f(x)=(x2−3)−2 with respect to *x*,

f′(x)=ddx((x2−3)−2)

Apply the power rule, difference rule and chain rule,

f′(x)=−2(x2−3)−2−1ddx(x2−3)=−2(x2−3)−3[ddx(x2)−ddx(3)]=−2(x2−3)−3[2x−0]=−4x(x2−3)−3

Therefore, the derivative of the function
f′(x)=−4x(x2−3)−3.

Substitute
f′(x)=−4x(x2−3)−3 in
dy=f′(x)dx,

dy=−4x(x2−3)−3dx=−4x(x2−3)3dx

Therefore, the differential of the function is
dy=−4x(x2−3)3dx.

#### To determine

**To find:** The differential of the function.

#### Answer

The differential of the function is
dy=−2t3dt1−t4.

#### Explanation

**Given:**

The function is
y=1−t4.

**Result used:**

If
y=f(x), then the differential is
dy=f′(x)dx.

**Calculation:**

The differential of the function is computed as follows,

Consider
y=f(t).

Differentiate
f(t)=1−t4 with respect to *t*,

f′(t)=ddt(1−t4)=ddt((1−t4)12)

Let
u=1−t4,

f′(t)=ddt((u)12)

Apply the chain rule (1) and simplify the terms,

f′(t)=ddu((u)12)dudt=(12u12−1)dudt=(12u−12)dudt=12ududt

Substitute
u=1−t4,

f′(t)=121−t4ddt(1−t4)=121−t4[ddt(1)−ddt(t4)]=121−t4[0−4t3]=−2t31−t4

Therefore, the derivative of the function
f′(t)=−2t31−t4.

Substitute
f′(t)=−2t31−t4 in
dy=f′(t)dt.

dy=(−2t31−t4)dt=−2t3dt1−t4

Therefore, the differential of the function is
dy=−2t3dt1−t4.