Differentiation

The derivative of $x^2$ is found using the power rule of differentiation, which states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$. Applying this rule, we get $f'(x) = 2x^{2-1} = 2x$.

To find the derivative of $sin(x)$ using the chain rule, we first identify the outer function $g(u) = sin(u)$ and the inner function $h(x) = x$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = cos(u)$ and $h'(x) = 1$. Plugging these values in, we get $f'(x) = cos(x) * 1 = cos(x)$.

The derivative of $e^x$ is found using the exponential rule of differentiation, which states that if $f(x) = e^x$, then $f'(x) = e^x$. Therefore, the derivative of $e^x$ is simply $e^x$.

The derivative of $ln(x)$ is found using the logarithmic rule of differentiation, which states that if $f(x) = ln(x)$, then $f'(x) = 1/x$. Therefore, the derivative of $ln(x)$ is simply $1/x$.

The derivative of a polynomial is found by applying the power rule of differentiation to each term. In this case, we have $f'(x) = d/dx(x^3) + d/dx(2x^2) + d/dx(-3x) + d/dx(4) = 3x^2 + 4x - 3$.

To find the derivative of $(x^2 + 1)(x - 3)$ using the product rule, we first identify the two functions $g(x) = x^2 + 1$ and $h(x) = x - 3$. Then, we apply the product rule formula: $f'(x) = g'(x) * h(x) + g(x) * h'(x)$. In this case, $g'(x) = 2x$ and $h'(x) = 1$. Plugging these values in, we get $f'(x) = 2x(x - 3) + (x^2 + 1)$.

To find the derivative of $\frac{x^2 + 1}{x - 3}$ using the quotient rule, we first identify the two functions $g(x) = x^2 + 1$ and $h(x) = x - 3$. Then, we apply the quotient rule formula: $f'(x) = \frac{g'(x)(x - 3) - g(x)h'(x)}{(x - 3)^2}$. In this case, $g'(x) = 2x$ and $h'(x) = 1$. Plugging these values in, we get $f'(x) = \frac{2x(x - 3) - (x^2 + 1)}{(x - 3)^2}$.

To find the derivative of $sin(3x + 2)$ using the chain rule, we first identify the outer function $g(u) = sin(u)$ and the inner function $h(x) = 3x + 2$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = cos(u)$ and $h'(x) = 3$. Plugging these values in, we get $f'(x) = cos(3x + 2) * 3 = 3cos(3x + 2)$.

To find the derivative of $e^(2x - 1)$ using the chain rule, we first identify the outer function $g(u) = e^u$ and the inner function $h(x) = 2x - 1$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = e^u$ and $h'(x) = 2$. Plugging these values in, we get $f'(x) = e^(2x - 1) * 2 = 2e^(2x - 1)$.

To find the derivative of $ln(x^2 + 1)$ using the chain rule, we first identify the outer function $g(u) = ln(u)$ and the inner function $h(x) = x^2 + 1$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = \frac{1}{u}$ and $h'(x) = 2x$. Plugging these values in, we get $f'(x) = \frac{1}{x^2 + 1} * 2x = \frac{2x}{x^2 + 1}$.

To find the derivative of $\frac{sin(x)}{cos(x)}$ using the quotient rule, we first identify the two functions $g(x) = sin(x)$ and $h(x) = cos(x)$. Then, we apply the quotient rule formula: $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}$. In this case, $g'(x) = cos(x)$ and $h'(x) = -sin(x)$. Plugging these values in, we get $f'(x) = \frac{cos(x)cos(x) - sin(x)(-sin(x))}{cos^2(x)} = \frac{cos^2(x) + sin^2(x)}{cos^2(x)} = \frac{1}{cos^2(x)} = \frac{cos(x) - sin(x)}{cos^2(x)}$.

To find the derivative of $tan(x)$ using the chain rule, we first identify the outer function $g(u) = tan(u)$ and the inner function $h(x) = x$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = sec^2(u)$ and $h'(x) = 1$. Plugging these values in, we get $f'(x) = sec^2(x) * 1 = sec^2(x)$.

To find the derivative of $cot(x)$ using the chain rule, we first identify the outer function $g(u) = cot(u)$ and the inner function $h(x) = x$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = -csc^2(u)$ and $h'(x) = 1$. Plugging these values in, we get $f'(x) = -csc^2(x) * 1 = -csc^2(x)$.

To find the derivative of $arcsin(x)$ using the chain rule, we first identify the outer function $g(u) = arcsin(u)$ and the inner function $h(x) = x$. Then, we apply the chain rule formula: $f'(x) = g'(h(x)) * h'(x)$. In this case, $g'(u) = \frac{1}{\sqrt{1 - u^2}}$ and $h'(x) = 1$. Plugging these values in, we get $f'(x) = \frac{1}{\sqrt{1 - x^2}} * 1 = \frac{1}{\sqrt{1 - x^2}}$.