Basic Rules of Differentiation

Approach: Use the sum and difference rule together with the power function derivative formula.

Steps:

1. Apply the sum and difference rule: $f'(x) = (x^3)' + (2x^2)' - (5x)' + (1)'$

2. Compute the derivative of each term: $(x^3)' = 3x^2$ $(2x^2)' = 2 \cdot 2x = 4x$ $(5x)' = 5$ $(1)' = 0$

3. Combine the results: $f'(x) = 3x^2 + 4x - 5$

Answer: $f'(x) = 3x^2 + 4x - 5$

Approach: Use the product rule.

Steps:

1. Let $u = x^2$ and $v = \sin x$

2. Compute $u'$ and $v'$: $u' = 2x$ $v' = \cos x$

3. Apply the product rule: $f'(x) = u'v + uv' = 2x \cdot \sin x + x^2 \cdot \cos x$

4. Simplify: $f'(x) = 2x\sin x + x^2\cos x$

Answer: $f'(x) = 2x\sin x + x^2\cos x$

Approach: Use the quotient rule.

Steps:

1. Let $u = x^2 + 1$ and $v = x - 1$

2. Compute $u'$ and $v'$: $u' = 2x$ $v' = 1$

3. Apply the quotient rule: $f'(x) = \frac{u'v - uv'}{v^2} = \frac{2x(x-1) - (x^2+1)(1)}{(x-1)^2}$

4. Expand: $= \frac{2x^2 - 2x - x^2 - 1}{(x-1)^2} = \frac{x^2 - 2x - 1}{(x-1)^2}$

Answer: $f'(x) = \frac{x^2 - 2x - 1}{(x-1)^2}$

Approach: Use the chain rule.

Steps:

1. Let $u = x^2 + 1$, so $f(x) = \sin u$

2. Compute the derivative of each layer: $\frac{df}{du} = \cos u = \cos(x^2 + 1)$ $\frac{du}{dx} = 2x$

3. Apply the chain rule: $f'(x) = \frac{df}{du} \cdot \frac{du}{dx} = \cos(x^2 + 1) \cdot 2x$

4. Simplify: $f'(x) = 2x\cos(x^2 + 1)$

Answer: $f'(x) = 2x\cos(x^2 + 1)$

Approach: Use the property of equivalent infinitesimals together with the definition of the derivative.

Steps:

1. From $\lim\limits_{x\to1} \frac{f(x)}{\ln x}=1$ we get $f(x)\sim \ln x$ as $x\to1$

2. As $x\to1$, $\ln x\to0$, so $\lim\limits_{x\to1} f(x)=0$

3. Let $g(x)=\frac{f(x)}{\ln x}$, then $f(x)=g(x)\ln x$

4. Apply the product rule: $f'(x)=g'(x)\ln x + g(x)\cdot\frac{1}{x}$

5. As $x\to1$, $\ln x\to0$, $\frac{1}{x}\to1$, $g(x)\to1$

6. Therefore $f'(1)=\lim\limits_{x\to1} f'(x)=1$

Answer: $f'(1)=1$

Approach: Use the parametric differentiation formula.

Steps:

1. When $t\geq0$, $x=3t, y=t\sin t$, so $y=\frac{x}{3}\sin\frac{x}{3}$

2. When $t<0$, $x=t, y=-t\sin t$, so $y=-x\sin x$

3. Hence $f(x)=\begin{cases}\frac{x}{3}\sin\frac{x}{3}, & x\geq0\\-x\sin x, & x<0\end{cases}$

4. Compute the left derivative: $f'_-(0)=\lim\limits_{x\to0^-}\frac{-x\sin x-0}{x}=-\lim\limits_{x\to0^-}\sin x=0$

5. Compute the right derivative: $f'_+(0)=\lim\limits_{x\to0^+}\frac{\frac{x}{3}\sin\frac{x}{3}-0}{x}=\lim\limits_{x\to0^+}\frac{\sin\frac{x}{3}}{3}=0$

6. Since the left and right derivatives are equal, $f'(0)=0$

Answer: $f'(0)=0$

Approach

This exercise tests the differentiation of a function with a variable upper limit: if $f(x) = \int_{a}^{x} g(t)\,dt$, then $f'(x) = g(x)$, which is a direct consequence of the fundamental theorem of calculus (the Newton–Leibniz formula). The plan is to identify the integrand first, then apply the formula; there is no need to find an antiderivative explicitly.

Steps

1. Identify the function with a variable upper limit: $f(x) = \int_0^x e^{\cos t}\,dt$, whose integrand is $g(t) = e^{\cos t}$ and whose upper limit is $x$.
2. By the fundamental theorem of calculus, $f'(x) = g(x) = e^{\cos x}$; the upper limit $x$ contributes a factor of $1$, and the constant lower limit $0$ produces no extra term.
3. Check: the integrand evaluated at $t = x$ is $e^{\cos x}$, so $f'(x) = e^{\cos x}$, which agrees with the result obtained by applying the formula directly.

Answer

$f'(x)=e^{\cos x}$

Approach

This exercise tests the differentiation of a function with a variable upper limit: if $f(x) = \int_{a}^{x} g(t)\,dt$, then $f'(x) = g(x)$, that is, “the derivative equals the value of the integrand at the upper limit”. Here the integrand itself has a composite structure, so the plan is to identify the integrand first and then apply the formula; there is no need to find an antiderivative explicitly.

Steps

1. Identify the function with a variable upper limit: $f(x) = \int_0^x e^{t^2} \sin t\,dt$, whose integrand is $g(t) = e^{t^2} \sin t$ and whose upper limit is $x$.
2. By the fundamental theorem of calculus, $f'(x) = g(x) = e^{x^2} \sin x$; the upper limit $x$ contributes a factor of $1$, and the constant lower limit $0$ produces no extra term.
3. Check: the integrand evaluated at $t = x$ is $e^{x^2} \sin x$, so $f'(x) = e^{x^2} \sin x$, which agrees with the result obtained by applying the formula directly.

Answer

$f'(x) = e^{x^2} \sin x$

Approach: Use the properties of the differential equation together with the definition of the derivative.

Steps:

1. Substitute $x=1$ into the differential equation: $y'(1)+\frac{1}{2\sqrt{1}}y(1)=2+\sqrt{1}$

2. Since $y(1)=3$: $y'(1)+\frac{1}{2}\cdot3=3$

3. Solving gives $y'(1)=3-\frac{3}{2}=\frac{3}{2}$

Answer: $y'(1)=\frac{3}{2}$

Approach: Use the properties of the function together with the definition of the derivative.

Steps:

1. From $f(x+2)-f(x)=x$, differentiate both sides with respect to $x$: $f'(x+2)-f'(x)=1$

2. When $x=0$: $f'(2)-f'(0)=1$

3. Since $f(x)$ is continuous and $\int_0^2 f(x)dx=0$, we can infer that $f(0)=0$

4. Use the definition of the derivative: $f'(0)=\lim\limits_{h\to0}\frac{f(h)-f(0)}{h}=\lim\limits_{h\to0}\frac{f(h)}{h}$

5. Since $f(x+2)-f(x)=x$, at $x=0$ we get $f(2)-f(0)=0$, so $f(2)=0$

6. Therefore $f'(0)=\frac{1}{2}$

Answer: $f'(0)=\frac{1}{2}$

Approach: Use Taylor’s formula together with the definition of the derivative.

Steps:

1. Expand to second order with Taylor’s formula: $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + o(x^2)$

2. Since $f(0)=0$ and $f'(0)=0$: $f(x) = \frac{f''(0)}{2}x^2 + o(x^2)$

3. Hence: $f''(0) = 2\lim\limits_{x\to0}\frac{f(x)}{x^2}$

4. Since $f(x)$ is continuous on $[-a,a]$ and $f(0)=0$, we get $f''(0)=0$

Answer: $f''(0)=0$

Approach: Use the derivative formula for a function with a variable upper limit.

Steps:

1. By the derivative formula for a function with a variable upper limit: $g'(x) = f(x,y)$

2. This is a property of probability density functions: the derivative of a function with a variable upper limit equals the value of the integrand at the upper limit

Answer: $g'(x) = f(x,y)$