Unit 2: Differential calculus and its applications - Subjective Questions

General rules:

• Product rule: If , then .
• Quotient rule: If , then , where .
• Chain rule: If , then .

For , apply the product and chain rules:

Now,

and

Therefore,

For a parametric curve,

Here,

Thus,

The second derivative is

Since

we obtain

Differentiate the equation with respect to :

Collecting the derivative terms gives

Hence,

Differentiate

once more:

Therefore,

so

where .

Logarithmic differentiation is useful when a variable occurs in both the base and exponent or when a function contains complicated products and quotients.

Take logarithms:

Differentiate implicitly:

Thus,

Substituting the original value of ,

Let

The first derivative is

Using the definitions of and ,

Therefore,

Every differentiation multiplies the amplitude by and increases the phase by . Repeating the process times gives

Equivalently,

Rolle's theorem: If a function is continuous on , differentiable on , and satisfies , then there exists at least one such that .

For :

• It is a polynomial, so it is continuous on and differentiable on .
• .
• .

Thus, all the hypotheses are satisfied. Now,

Setting gives

Since , Rolle's theorem is verified, with

Lagrange's mean value theorem: If is continuous on and differentiable on , then there exists at least one such that

The function is continuous on and differentiable on . Its average rate of change is

Also,

Hence,

Therefore,

Since , the required value is

Cauchy's mean value theorem: Let and be continuous on and differentiable on . If on , then there exists at least one such that

Equivalently,

Comparison:

• Cauchy's theorem involves two functions, and .
• Lagrange's theorem involves one function and states
  
• Lagrange's theorem is a special case of Cauchy's theorem obtained by taking .
• Both theorems require continuity on the closed interval and differentiability on the open interval.
• Cauchy's theorem is important in proving results such as L'Hospital's rule.

Taylor's theorem: If has the required derivatives near , then

where the Lagrange remainder is

for some between and .

For about :

Therefore,

Since

the remainder is

where lies between and . Hence,

The standard Maclaurin series are




To expand , multiply

Collecting terms through degree four gives

Thus,

The coefficient of is zero.

An indeterminate form is a symbolic form obtained during limit evaluation that does not uniquely determine the value of the limit.

The standard indeterminate forms are

Transformations:

• Forms and may be treated directly using algebraic simplification or L'Hospital's rule.
• A product of the form can be rewritten as
  
• A difference of the form can often be handled by combining fractions, rationalizing, or factoring.
• For exponential forms, set and take logarithms:
  
  Evaluate the limit of first and then exponentiate to obtain the limit of .

L'Hospital's rule: If has the form or and the required derivatives exist, then

provided the limit on the right exists.

(i) Applying the rule repeatedly,

(ii)

(iii) Let . Then

Using L'Hospital's rule,

Therefore,

Differentiate the function:

The stationary points satisfy , so

The second derivative is

At ,

so is a point of local maximum. Its value is

At ,

so is a point of local minimum. Its value is

Therefore:

• Local maximum: at .
• Local minimum: at .

Let the side of the square base be and the height be .

The volume constraint is

so

The surface area of the open box is

Substituting for ,

Differentiate:

For a stationary point,

Thus,

The height is

Also,

for , confirming a minimum. Therefore, the required dimensions are

Derivative of : Let

Differentiating,

Therefore,

Since ,

Derivative of : Let

Differentiating,

Hence,

Thus,

The other standard derivatives are

Leibniz's theorem: If and are sufficiently differentiable, then

Take

The nonzero derivatives of are

while every derivative of is . Therefore, only the terms for remain:

Since

we obtain

The Maclaurin expansion is

valid for .

Set :

Thus,

This is an alternating series whose term magnitudes decrease. Therefore, the absolute error is no greater than the magnitude of the first omitted term:

Hence,

For the curve,

At ,

The coordinates of the point are

Tangent: Its slope is , so

Therefore,

Normal: Its slope is the negative reciprocal of , namely . Thus,

which simplifies to

Differentiate

with respect to :

Hence,

Differentiate again:

Using the quotient rule,

Substituting gives

Since ,

Closed-interval procedure:

• Find all critical points in the open interval where or does not exist.
• Evaluate at every critical point.
• Evaluate at both endpoints.
• Compare all values; the largest is the absolute maximum and the smallest is the absolute minimum.

For

we have

Thus, the critical points in are and .

Evaluate the function:




Comparing these values:

• The absolute maximum is at .
• The absolute minimum is at both and .