For the parametric curve \( x = t^{2} \), \( y = t^{3} \), the slope \( dy/dx \) is
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
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Logarithmic differentiation is the natural method for a function of the form
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
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For \( y = x^{x} \) with \( x > 0 \), the derivative is
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
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The tangent to \( y = x^{3} - 3x \) is horizontal at
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
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A function is increasing on an interval when, throughout it,
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
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A point of inflexion on a twice-differentiable curve is one where
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
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The point of inflexion of \( y = x^{3} - 6x^{2} + 9x \) lies at
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
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The Taylor expansion of \( f(x,y) \) about \( (a,b) \) has first-order terms
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
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Three of these four appear in the second-order group of a two-variable Taylor expansion. The exception is
Answer: A
The second-order group is \( \tfrac{1}{2}\left[h^{2}f_{xx} + 2hk\,f_{xy} + k^{2}f_{yy}\right] \) with \( h = x-a \) and \( k = y-b \). A product of two *first* derivatives never appears: the expansion is built from single derivatives of increasing order, not from products of lower ones
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Match the failure of continuity with its description. 1. Removable 2. Jump 3. Infinite P. The one-sided limits exist and differ Q. The limit exists but the value assigned differs R. The function grows without bound at the point
Answer: D
A removable discontinuity is repaired by reassigning one value, the limit already existing. A jump has two one-sided limits that disagree, so no reassignment helps. An infinite discontinuity has no finite limit on at least one side, the function running away as the point is approached
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