Questions and Answers
What is the derivative of the function $f(x) = e^{x^2}$ with respect to $x$?
Which of the following functions has the derivative $f'(x) = 6x^2 - 4$?
For the function $g(x) = an(x^2)$, what is the second derivative $g''(x)$?
What is the value of the limit $\lim_{x \to 0} \frac{sin(5x)}{x}$?
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Determine the critical points of the function $h(x) = x^4 - 8x^2 + 16$.
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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!
