# Can anyone explain the concept of derivatives?

Aug 26, 2016

This is a very vast question. Honestly, it's hard to explain in a few words.

By definition, the derivative of the function is the instantaneous rate of change with respect to $x$ at any point on that function. In other words, the derivative measures the slope at any given point in the domain of the function.

There are many different techniques to find derivatives of functions, as well as vast applications. Here are a few examples:

Example 1: Determine the derivative of $y = 3 {x}^{4} \times 3 {x}^{7}$

$y = 9 {x}^{11}$

By the power rule:

$y ' = 99 {x}^{10}$

Example 2: Determine the derivative of $y = 2 \sqrt{x + 8}$

First, let's differentiate the sqrt using the chain rule.

Let $y = {u}^{\frac{1}{2}}$ and $u = x + 8$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{2} {u}^{- \frac{1}{2}} \times 1 = \frac{1}{2 {\left(x + 8\right)}^{\frac{1}{2}}} = \frac{1}{2 \sqrt{x + 8}}$

Now, we can calculate the derivative of the entire function using the product rule.

$y ' = 0 \times \sqrt{x + 8} + 2 \times \frac{1}{2 \sqrt{x + 8}}$

$y ' = \frac{1}{\sqrt{x + 8}}$

Example 3: Determine the derivative of the relation ${x}^{2} {y}^{2} - {x}^{3} y = x y$

Through implicit differentiation and the product rule, we get:

${x}^{2} {y}^{2} - {x}^{3} y - x y = 0$

$2 x {y}^{2} + 2 {x}^{2} y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - 3 {x}^{2} y - {x}^{3} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - y - x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

$2 {x}^{2} y \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - {x}^{3} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) - x \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = 3 {x}^{2} y + y - 2 x {y}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 {x}^{2} y - {x}^{3} - x\right) = 3 {x}^{2} y + y - 2 x {y}^{2}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3 {x}^{2} y + y - 2 x {y}^{2}}{2 {x}^{2} y - {x}^{3} - x}$

Example 4: Determine the derivative of the following: $y = {2}^{\cos} x$

Taking the natural logarithm of both sides:

$\ln y = \ln \left({2}^{\cos x}\right)$

$\ln y = \cos x \ln 2$

Differentiate the right side using the product rule and the left side using the derivative rule $\left(\ln x\right) ' = \frac{1}{x}$:

$\frac{1}{y} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = - \sin x \times \ln 2 + 0 \times \cos x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- \ln 2 \sin x}{\frac{1}{y}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \times - \ln 2 \sin x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {2}^{\cos} x \times - \ln 2 \sin x$

Example 5: Determine the equation of the tangent line to the curve $y = \csc x$ at the point $x = \frac{11 \pi}{6}$

The tangent line to the curve means the line that touches the curve only once, at the given point.

We need to start by finding the point of tangency. This is done by evaluating $x = a$, where $a$ is the given point, into the function.

$y = \csc \left(\frac{11 \pi}{6}\right)$

$y = \frac{1}{\sin \left(\frac{11 \pi}{6}\right)}$

$y = \frac{1}{- \frac{1}{2}}$

$y = - 2$

Next, we need to differentiate. The derivative of $y = \csc x$ is given by the quotient rule:

$y = \csc x$

$y = \frac{1}{\sin} x$

$y ' = \frac{0 \times \sin x - 1 \times \cos x}{\sin x} ^ 2$

$y ' = - \cos \frac{x}{\sin} ^ 2 x$

$y ' = - \left(\cos \frac{x}{\sin} x\right) \left(\frac{1}{\sin} x\right)$

$y ' = - \cot x \csc x$

Next, we need to determine the slope of the tangent. This can be found by evaluating $x = a$ into the derivative.

$y = - \cot \left(\frac{11 \pi}{6}\right) \times \csc \left(\frac{11 \pi}{6}\right)$

$y = - \left(- \sqrt{3}\right) \times - 2 = - 2 \sqrt{3}$

The last step to problems such as these is to determine the equation of the line using point-slope form.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

y - (-2) = -2sqrt(3)(x - (11pi)/6))

$y + 2 = - 2 \sqrt{3} x + \frac{11 \pi \sqrt{3}}{3}$

$y = - 2 \sqrt{3} x + \frac{11 \pi \sqrt{3} - 6}{3}$

There are other applications to derivatives, but this answer is already extremely long and so I'll leave it up to other contributors to explain curve sketching and applications involving optimization, biological sciences, economics and geometric figures.

Hopefully this overview has enlightened you towards calculus and has been helpful.

Aug 26, 2016

Here's a visual way to approach explaining derivatives.

A derivative is the instantaneous rate of change, like at a specific point on a graph. So, it is the slope over very small changes in $y$ and $x$.

Mathematically, you have:

${\lim}_{\Delta x \to 0} \frac{f \left(x + \Delta x\right) - f \left(x\right)}{\Delta x}$

This is really saying the same thing as $\frac{\Delta \left(f \left(x\right)\right)}{\Delta x}$ or $\frac{\Delta y}{\Delta x}$ for small $\Delta x$ (and consequently $\Delta \left(f \left(x\right)\right)$).

Conceptually, take a graph on a calculator, and zoom in.

• Take the idea of the slope, $\frac{\Delta y}{\Delta x}$ (rise over run), and consider really small $\Delta x$ and $\Delta y$.
• You can achieve small values of $\Delta x$ and $\Delta y$ by zooming into a graph really closely, since your viewing window is progressively looking at smaller and smaller regions.

For example, velocity is the derivative of position, so zooming into a position vs. time graph gives you the instantaneous velocity within the small interval $\left[a , b\right]$: When your graph looks linear at your specific spot on your graph, that represents what it means to take the derivative at that spot.