Set of ten calculus exercises featuring limits, derivatives, and integration problems for students.

Let's solve each of the 10 problems step by step.

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

Step 1: Recognize 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 as $x \to 7$:
$$
\lim_{x \to 7} (x + 7) = 7 + 7 = \boxed{14}
$$

✔ Answer: 14

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

This is a standard limit:

$$
\lim_{x \to 0} \frac{\sin x}{x} = \boxed{1}
$$

✔ Answer: 1

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

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

$$
f'(x) = 2 \cdot 5x^4 + 6 \cdot 3x^2 - 3 \cdot 1 + 0
$$
$$
= 10x^4 + 18x^2 - 3
$$

✔ Answer: $f'(x) = 10x^4 + 18x^2 - 3$

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

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

Let $u = \cos x$, $v = \sin x$

Then:
- $u' = -\sin x$
- $v' = \cos x$

So:
$$
\frac{d}{dx}(\cos x \sin x) = (-\sin x)(\sin x) + (\cos x)(\cos x)
= -\sin^2 x + \cos^2 x
= \cos^2 x - \sin^2 x
$$

Alternatively, recall: $\cos(2x) = \cos^2 x - \sin^2 x$

So this is also equal to $\cos(2x)$

✔ Answer: $\cos^2 x - \sin^2 x$ or $\boxed{\cos(2x)}$

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

So:
$$
\int_2^3 (6x^2 + 5)\,dx = 69 - 26 = \boxed{43}
$$

✔ Answer: 43

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

Note: $\log^2(x)$ means $(\log x)^2$, not $\log(\log x)$

Let $u = \log x$, so $y = u^2$

Then:
$$
\frac{dy}{dx} = 2u \cdot \frac{du}{dx} = 2 \log x \cdot \frac{1}{x}
$$

So:
$$
\frac{d}{dx}(\log^2 x) = \boxed{\frac{2 \log x}{x}}
$$

✔ Answer: $\frac{2 \log x}{x}$

*(Assuming $\log x = \ln x$, natural log — common in calculus)*

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

Factor out constant:
$$
= -20 \int_1^3 x^3\,dx = -20 \left[ \frac{x^4}{4} \right]_1^3
= -20 \left( \frac{81}{4} - \frac{1}{4} \right)
= -20 \left( \frac{80}{4} \right)
= -20 \cdot 20 = \boxed{-400}
$$

✔ Answer: -400

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

First derivative:
$$
\frac{d}{dx}(x^3 - x + 7) = 3x^2 - 1
$$

Second derivative:
$$
\frac{d}{dx}(3x^2 - 1) = 6x
$$

✔ Answer: $\boxed{6x}$

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9) $f(x) = 2x^5 - 3x$, find $f'(2)$

First, find $f'(x)$:

$$
f'(x) = 2 \cdot 5x^4 - 3 = 10x^4 - 3
$$

Now plug in $x = 2$:
$$
f'(2) = 10(16) - 3 = 160 - 3 = \boxed{157}
$$

✔ Answer: 157

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

Integrate term by term:

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

Now 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

So:
$$
\int_0^2 = 58 - 0 = \boxed{58}
$$

✔ Answer: 58

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✔ Final Answers Summary:

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

Let me know if you'd like any explanation clarified!