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Advanced Calculus Derivatives Quiz

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Questions and Answers

What is the derivative of the function $f(x) = e^{x^2}$ with respect to $x$?

$2xe^{x^2}$ (correct)

$xe^{x^2}$

$e^{x^2} + 2xe^{x^2}$

$e^{x^2} - 2x^2e^{x^2}$

Which of the following functions has the derivative $f'(x) = 6x^2 - 4$?

$3x^3 - 4x + C$

$2x^3 + 4x + C$

$2x^3 - 4x + C$ (correct)

$x^3 - 4x + C$

For the function $g(x) = an(x^2)$, what is the second derivative $g''(x)$?

$2 ext{sec}^2(x^2)(1 + 2x ext{tan}(x^2))$ (correct)

$4x ext{sec}^2(x^2)$

$2 an(x^2) ext{sec}^2(x^2)$

$ an(x^2)(4 + 2 an(x^2))$

What is the value of the limit $\lim_{x \to 0} \frac{sin(5x)}{x}$?

<p>$5$</p> Signup and view all the answers

Determine the critical points of the function $h(x) = x^4 - 8x^2 + 16$.

<p>$x = 2$ and $x = -2$</p> Signup and view all the answers

Study Notes

Derivatives and Limits

The derivative of the function ( f(x) = e^{x^2} ) with respect to ( x ) is ( f'(x) = 2xe^{x^2} ) using the chain rule.
A function with the derivative ( f'(x) = 6x^2 - 4 ) can be determined by integrating to find ( f(x) = 2x^3 - 4x + C ), where ( C ) is a constant.

Second Derivative

For the function ( g(x) = \tan(x^2) ), the first derivative is ( g'(x) = 2x \sec^2(x^2) ).
The second derivative ( g''(x) ) involves applying the product and chain rules, yielding ( g''(x) = 2\sec^2(x^2) + 8x^2\tan(x^2)\sec^2(x^2) ).

Limit Evaluation

The limit ( \lim_{x \to 0} \frac{\sin(5x)}{x} ) evaluates to ( 5 ) using the fact that ( \lim_{x \to 0} \frac{\sin(kx)}{x} = k ).

Critical Points

The critical points of the function ( h(x) = x^4 - 8x^2 + 16 ) are found by calculating the derivative ( h'(x) = 4x^3 - 16x ) and setting it to zero.
Solving ( 4x(x^2 - 4) = 0 ) gives critical points at ( x = 0, x = -2, x = 2 ).

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Description

Challenge your understanding of derivatives and differentiation in this advanced calculus quiz. Solve intricate problems involving derivatives, critical points, and limits. Perfect for students looking to deepen their calculus skills!