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Differential Calculus Questions and Answers

51.

Consider the following statements about differentiability. 1. A function differentiable at a point is continuous there   2. A function continuous on a closed interval attains its bounds   3. A function with a corner has a derivative at the corner

Answer: C

The first is the standard implication, differentiability being the stronger property. The second is the extreme value theorem, which needs both closedness and continuity. The third is false: at a corner the one-sided derivatives exist but disagree, as they do for \( |x| \) at the origin, so no single derivative exists

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52.

Consider the following statements about partial derivatives. 1. The mixed derivatives commute whenever they are continuous   2. Euler's theorem weights each first derivative by its own variable   3. A vanishing Jacobian means the functions are independent

Answer: B

Clairaut's theorem gives the first, and Euler's theorem \( x f_{x} + y f_{y} = n f \) the second. The third reverses the truth: a vanishing Jacobian marks functional *dependence*, a relation \( \phi(u,v) = 0 \) tying the two together, and independence is what a non-zero Jacobian certifies

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53.

Put these curves in order of increasing radius of curvature at the origin. 1. \( y = x^{2} \)   2. \( y = 2x^{2} \)   3. \( y = x \)   4. \( y = 4x^{2} \)

Answer: A

For \( y = ax^{2} \) the slope vanishes at the origin and \( y'' = 2a \), so

\[ \rho \;=\; \frac{1}{2a}, \]

giving \( 1/8 \), \( 1/4 \) and \( 1/2 \) for \( a = 4, 2, 1 \). The straight line has zero curvature and so infinite radius, placing it last

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Answer: D

The theorem is an existence result: it promises some \( c \) with \( f'(c) \) equal to the average slope, without saying where. Finding \( c \) means solving an equation, which for most functions cannot be done in closed form. The two hypotheses and the existence conclusion are all stated correctly

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Answer: C

The difference quotient \( [f(x)-f(a)]/(x-a) \) can only tend to a finite limit if the numerator tends to zero, which is continuity at \( a \). So differentiability is strictly the stronger condition, and a discontinuity of any kind, removable or not, rules the derivative out

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Answer: B

The condition is necessary for an interior extremum of a differentiable function but not sufficient. The function \( y = x^{3} \) has \( y' = 0 \) at the origin yet rises through it, the point being a horizontal inflexion. Deciding which case holds needs the second derivative, or the first non-vanishing one

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57.

If \( y = x^{3} \) and \( x = 2 \) is measured with an error of 1 per cent, the resulting percentage error in \( y \) is about

Answer: A

Take logarithms and differentiate, which converts a product into a sum of relative errors:

\[ \ln y = 3 \ln x \quad \Longrightarrow \quad \frac{dy}{y} \;=\; 3\,\frac{dx}{x}. \]

So a one per cent error in \( x \) becomes three per cent in \( y \). The exponent multiplies the relative error, and the value \( x = 2 \) does not enter at all

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Answer: D

The differential is the linear part of the increment, \( dy = f'(x)\,dx \), and it approximates the true increment \( \Delta y \) with an error of order \( dx^{2} \). The bare difference of two values of \\( f \\) is that exact increment rather than its linear approximation, and the version keeping a term in \\( dx^{2} \\) carries a second-order term the differential deliberately omits

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59.

Using differentials, \( \sqrt{16.1} \) is approximately

Answer: C

Take \( y = \sqrt{x} \) at \( x = 16 \), where \( y = 4 \) and \( dy/dx = 1/(2\sqrt{x}) = 1/8 \). With \( dx = 0.1 \),

\[ \sqrt{16.1} \;\approx\; 4 + \frac{1}{8}(0.1) \;=\; 4 + 0.0125 \;=\; 4.0125. \]

The true value is 4.01248, so the linear estimate is high by two parts in a million

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60.

For the circle \( x^{2} + y^{2} = 25 \), the slope \( dy/dx \) at the point \( (3, 4) \) is

Answer: B

Differentiate implicitly and solve for the slope:

\[ 2x + 2y\,\frac{dy}{dx} \;=\; 0 \quad \Longrightarrow \quad \frac{dy}{dx} \;=\; -\frac{x}{y}. \]

At \( (3,4) \) this is \( -3/4 \). The tangent is perpendicular to the radius, whose slope is \( 4/3 \), and the product of the two slopes is \( -1 \) as it must be

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