Calculus practice problems covering limits, derivatives, and integrals.

Let’s solve each problem step by step with clear explanations.

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1) \(\lim_{x \to 7} \frac{x^2 - 49}{x - 7}\)

Note that \(x^2 - 49 = (x - 7)(x + 7)\), so:

\[
\frac{x^2 - 49}{x - 7} = \frac{(x - 7)(x + 7)}{x - 7}
\]

For \(x \ne 7\), we can cancel \(x - 7\):

\[
= x + 7
\]

Now take the limit:

\[
\lim_{x \to 7} (x + 7) = 7 + 7 = 14
\]

✔ Answer: 14

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2) \(\lim_{x \to 0} \frac{\sin x}{x}\)

This is a standard limit in calculus. It is known that:

\[
\lim_{x \to 0} \frac{\sin x}{x} = 1
\]

You can prove this using the Squeeze Theorem or Taylor series, but it’s usually taken as a given.

✔ Answer: 1

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3) Let \(f(x) = 2x^5 + 6x^3 - 3x + 7\). Find the first derivative.

Use the power rule: \(\frac{d}{dx}(x^n) = n x^{n-1}\)

\[
f'(x) = \frac{d}{dx}(2x^5) + \frac{d}{dx}(6x^3) - \frac{d}{dx}(3x) + \frac{d}{dx}(7)
\]

\[
= 2 \cdot 5x^4 + 6 \cdot 3x^2 - 3 \cdot 1 + 0
\]

\[
= 10x^4 + 18x^2 - 3
\]

✔ Answer: \(f'(x) = 10x^4 + 18x^2 - 3\)

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4) \(\frac{d}{dx}(\cos x \sin x)\)

Use the product rule: \((uv)' = u'v + uv'\)

Let \(u = \cos x\), \(v = \sin x\)

Then \(u' = -\sin x\), \(v' = \cos x\)

So:

\[
\frac{d}{dx}(\cos x \sin x) = (-\sin x)(\sin x) + (\cos x)(\cos x)
\]

\[
= -\sin^2 x + \cos^2 x = \cos(2x)
\]

*(since \(\cos(2x) = \cos^2 x - \sin^2 x\))*

✔ Answer: \(\cos(2x)\)

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5) \(\int_2^3 (6x^2 + 5) dx\)

Integrate term by term:

\[
\int (6x^2 + 5) dx = 6 \cdot \frac{x^3}{3} + 5x = 2x^3 + 5x
\]

Evaluate from 2 to 3:

At \(x = 3\): \(2(27) + 5(3) = 54 + 15 = 69\)

At \(x = 2\): \(2(8) + 5(2) = 16 + 10 = 26\)

Subtract:

\[
69 - 26 = 43
\]

✔ Answer: 43

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6) \(\frac{d}{dx}(\log^2(x))\)

Assuming \(\log^2(x) = (\log x)^2\) (common notation for square of log, not log of log).

Let \(u = \log x\), then \(y = u^2\)

By chain rule:

\[
\frac{dy}{dx} = 2u \cdot \frac{du}{dx} = 2 \log x \cdot \frac{1}{x}
\]

✔ Answer: \(\frac{2 \log x}{x}\)

*(Note: If base is not specified, it's often natural log (ln), but same derivative form applies if base is constant.)*

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7) \(\int_1^3 (-20x^3) dx\)

Factor out constant:

\[
-20 \int_1^3 x^3 dx = -20 \cdot \left[ \frac{x^4}{4} \right]_1^3
\]

Compute:

At \(x = 3\): \(\frac{81}{4}\)

At \(x = 1\): \(\frac{1}{4}\)

Difference: \(\frac{81}{4} - \frac{1}{4} = \frac{80}{4} = 20\)

Multiply by -20:

\[
-20 \cdot 20 = -400
\]

✔ Answer: -400

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8) \(\frac{d^2}{dx^2}(x^3 - x + 7) dx\)

Wait — this notation is slightly off. It should be:

> \(\frac{d^2}{dx^2}(x^3 - x + 7)\)

That is, find the second derivative of the function.

First derivative:

\[
\frac{d}{dx}(x^3 - x + 7) = 3x^2 - 1
\]

Second derivative:

\[
\frac{d}{dx}(3x^2 - 1) = 6x
\]

✔ Answer: \(6x\)

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9) \(f(x) = 2x^5 - 3x\). Find \(f'(2)\)

First, find \(f'(x)\):

\[
f'(x) = 10x^4 - 3
\]

Now plug in \(x = 2\):

\[
f'(2) = 10(16) - 3 = 160 - 3 = 157
\]

✔ Answer: 157

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10) \(\int_0^2 (8x^3 + 9x^2 - 2x + 3) dx\)

Integrate term by term:

\[
\int (8x^3 + 9x^2 - 2x + 3) dx = 8 \cdot \frac{x^4}{4} + 9 \cdot \frac{x^3}{3} - 2 \cdot \frac{x^2}{2} + 3x
\]

Simplify:

\[
= 2x^4 + 3x^3 - x^2 + 3x
\]

Evaluate from 0 to 2:

At \(x = 2\):

\[
2(16) + 3(8) - (4) + 3(2) = 32 + 24 - 4 + 6 = 58
\]

At \(x = 0\): all terms are 0 → 0

So integral = \(58 - 0 = 58\)

✔ Answer: 58

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## ✔ Final Answers Summary:

1) 14
2) 1
3) \(10x^4 + 18x^2 - 3\)
4) \(\cos(2x)\)
5) 43
6) \(\frac{2 \log x}{x}\)
7) -400
8) \(6x\)
9) 157
10) 58

Let me know if you want any step explained further!