Question

Double differentiate the following :
$\frac{4}{ (x-2t)^{2} + 1 }$
i) with respect to t (taking x as a constant)
ii) with respect to x ( taking t as a constant)

Answer 1

The two first derivatives differ by a factor of -2. The 2nd derivatives differ by a factor of (-2)×(-2)= 4. So we can do derivatives wrt x and write derivatives wrt t directly.

1)
$\frac{ {d}^{2}y }{d {t}^{2} } at \: x \: const = 32 \: \frac{3 \: {(x - 2t)}^{2} - 1 }{ {((x - 2t)^{2} + 1) }^{3} }$
2)
$\frac{ {d}^{2}y }{d {x}^{2} } at \: t \: const = 8 \: \frac{3 \: {(x - 2t)}^{2} - 1 }{ {((x - 2t)^{2} + 1) }^{3} }$
See the two enclosed pictures for details.

Answer 2

Answer:

The terminology "with respect to x" just means that x is the variable that we are changing, and we want to see how y reacts to changes in x. ... Well, we would use implicit differentiation again, except the variable of differentiation would be x, not y.

Step-by-step explanation:

Determining the derivative of a function from first principles requires a long calculation and it is easy to make mistakes. However, we can use this method of finding the derivative from first principles to obtain rules which make finding the derivative of a function much simpler.

Rules for differentiation

General rule for differentiation:

\[\frac{d}{dx}\left[{x}^{n}\right]=n{x}^{n-1}, \text{ where } n \in \mathbb{R} \text{ and } n \ne 0.\]

The derivative of a constant is equal to zero.

\[\frac{d}{dx}\left[k\right]= 0\]

The derivative of a constant multiplied by a function is equal to the constant multiplied by the derivative of the function.

\[\frac{d}{dx}\left[k \cdot f\left(x\right) \right]=k \frac{d}{dx}\left[ f\left(x\right) \right]\]

The derivative of a sum is equal to the sum of the derivatives.

\[\frac{d}{dx}\left[f\left(x\right)+g\left(x\right)\right]=\frac{d}{dx}\left[f\left(x\right) \right] + \frac{d}{dx}\left[g\left(x\right)\right]\]

The derivative of a difference is equal to the difference of the derivatives.

\[\frac{d}{dx}\left[f\left(x\right) - g\left(x\right)\right]=\frac{d}{dx}\left[f\left(x\right) \right] - \frac{d}{dx}\left[g\left(x\right)\right]\]