Calculus I, Fall 2014

Brief Answers to Quiz 2

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1. Evaluate \frac{d^{4}}{dx^{4}} (x^{4} cos(x − 1))|_{x=1}.

Ans:

\frac{d^{4}}{dx^{4}} (x^{4} cos(x − 1))|_{x=1} = [24 cos(x − 1) − 96x sin(x − 1) − 72x^{2} cos(x − 1)
+ 16x^{3} sin(x − 1) + x^{4} cos(x − 1)]|_{x=1}
= −47

2. Find the derivative of y = tan(exp(\sqrt{x^{2} + 1})). Need not simplify your final expression. Ans:

\frac{dy}{dx} = sec^{2}(exp(\sqrt{x^{2} + 1})) × exp(\sqrt{x^{2} + 1}) × \frac{x}{\sqrt{x^{2} + 1}}

3. Suppose we know that \frac{d}{dx}x^{n} = nx^{n−1} for all integers n. Show that, based on this fact, we can derive the same for n = q/p where p, q are integers and p ≠ 0. Ans: Let

y = x^{\frac{q}{p}} ⟹ y^{p} = x^{q}

Implicit differentiation gives us

py^{p−1} \frac{dy}{dx} = qx^{q−1}

hence

\frac{d}{dx}x^{\frac{q}{p}} = \frac{dy}{dx} = \frac{q}{p} \frac{x^{q−1}}{y^{p−1}} = \frac{q}{p} \frac{x^{q−1}}{(x^{\frac{q}{p}})^{p−1}} = \frac{q}{p}x^{q−1− \frac{q}{p} (p−1)} = \frac{q}{p}x^{\frac{q}{p} −1}

that is

4. Use implicit differentiation (and not other methods) to find dy/dx and d^{2}y/dx^{2} at (1, 1) where y(x) is implicitly given by x^{4} + y^{4} = 2. Ans: Using Implicit differentiation, we have

4x^{3} + 4y^{3} \frac{dy}{dx} = 0

substitute (x, y) = (1, 1) can get

4 + 4\frac{dy}{dx}|_{(1,1)} = 0 ⟹ \frac{dy}{dx}|_{(1,1)} = −1

using Implicit differentiation again to get

12x^{2} + 12y^{2}(\frac{dy}{dx})^{2} + 4y^{3} \frac{d^{2}y}{dx^{2}} = 0

substitute (x, y) = (1, 1) can get

12 + 12(−1)^{2} + 4\frac{d^{2}y}{dx^{2}} |_{(1,1)} = 0 ⟹ \frac{d^{2}y}{dx^{2}} |_{(1,1)} = −6

5. True or False? (prove it if true, correct it if false). If f, g and h are differentiable functions on R and f(g(x)) = h(x). Let \frac{d}{dx}f(x) = f_{1}(x), \frac{d}{dx}g(x) = g_{1}(x), \frac{d}{dx}h(x) = h_{1}(x). Then f_{1}(x) · g_{1}(x) = h_{1}(x). Ans: False. Correct version:

f_{1}(g(x)) · g_{1}(x) = h_{1}(x)