Differential Calculus Questions and Answers

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

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

Answer: A

Divide the two parametric derivatives:

\[ \frac{dy}{dx} \;=\; \frac{dy/dt}{dx/dt} \;=\; \frac{3t^{2}}{2t} \;=\; \frac{3t}{2}. \]

The slope vanishes at \( t = 0 \), where the curve has a cusp rather than a smooth horizontal tangent, since both parametric derivatives vanish together there

Answer: D

When both the base and the exponent vary, neither the power rule nor the exponential rule applies. Taking logarithms gives \( \ln y = v \ln u \), which differentiates by the product rule, and multiplying back by \( y \) recovers the derivative. Sums, quotients of polynomials and composed trigonometric functions all yield to the ordinary rules

Answer: C

Take logarithms, then differentiate both sides:

\[ \ln y = x \ln x \quad \Longrightarrow \quad \frac{1}{y}\frac{dy}{dx} \;=\; \ln x + 1. \]

So \( dy/dx = x^{x}(1 + \ln x) \), which vanishes at \( x = 1/e \), the minimum of the function. Applying the power rule as though the exponent were constant gives \\( x\\,x^{x-1} \\) instead

Answer: B

Set the derivative to zero:

\[ \frac{dy}{dx} \;=\; 3x^{2} - 3 \;=\; 0 \quad \Longrightarrow \quad x^{2} = 1. \]

So \( x = \pm 1 \), a maximum at \( -1 \) and a minimum at \( +1 \). The points \( \pm\sqrt{3} \) are where the curve meets the axis, and the origin is its inflexion

Answer: A

A positive first derivative means the function rises as \( x \) rises, which is what increasing means. A positive second derivative describes convexity, the curve bending upwards, and says nothing about direction: \( y = x^{2} \) is convex everywhere yet decreasing on the negative axis. The sign of the function itself is unrelated to either

Answer: D

An inflexion is a change of the sense of bending, so the second derivative must actually change sign, not merely vanish. For \( y = x^{4} \) the second derivative is zero at the origin but stays positive on both sides, and the point is a minimum. A change of sign of the first derivative marks an extremum instead

Answer: C

Differentiate twice and set the result to zero:

\[ \frac{dy}{dx} = 3x^{2} - 12x + 9, \qquad \frac{d^{2}y}{dx^{2}} = 6x - 12 \;=\; 0. \]

So \( x = 2 \), and the second derivative changes from negative to positive there, confirming a genuine inflexion. The values \( x = 1 \) and \( x = 3 \) are the stationary points

Answer: B

Each first derivative is weighted by the increment of its own variable, so the linear part is

\[ f(a,b) + (x-a)f_{x}(a,b) + (y-b)f_{y}(a,b). \]

This is the tangent plane to the surface at the point. The pairing that crosses the two derivatives attaches each to the wrong increment, and the term in \\( f_{xy} \\) belongs to the second-order group