Differentiation and Derivatives

The derivative of a function is the rate of change of the function with respect to its input. It is defined as the limit of the difference quotient as the change in the input approaches zero.

The derivative of a complex function (f(z)) is defined as (f'(z) = \lim_{h\to 0} \frac{f(z+h) - f(z)}{h}). Using this definition, we can find that the derivative of (f(z) = z^2) is (f'(z) = 2z). Therefore, the derivative of (f(z) = z^2) at (z = 2) is (f'(2) = 2(2) = 4).

A derivative is the rate of change of a function with respect to its input.

The derivative of the sine function is the cosine function, meaning $$d/dx(sin(x)) = cos(x)$$.

The formula for Nilakantha's method for finding the derivative of a function is $$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x - h)}{2h}$$. This formula is very similar to the formula for the derivative of a function that is used in modern calculus.

To find the derivative, we can use the power rule of differentiation. The derivative of $x^n$ is $n x^{n-1}$. Applying this rule, we get $f'(x) = \frac{d}{dx} (x^3 - 2x^2 + 3x - 4) = 3x^2 - 4x + 3$.

The derivative of (f(x)) can be found using the power rule of differentiation: (f'(x) = \frac{d}{dx}(x^3 - 2x^2 + 3x - 5) = 3x^2 - 4x + 3).

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 $f'(x) = cos(3x + 2) * 3 = 3cos(3x + 2)$.