Find the differential of each function. (a) y=tan (7t), (b) y=3-v^2/3+v^2

The main purpose of this question is to find the differential of each given function.

A function is a fundamental mathematical concept that describes a relationship between a set of inputs and a set of possible outputs, with each input corresponding to one output. The input is an independent variable and the output is referred to as a dependent variable.

Differential calculus and  integral calculus are the fundamental classifications of calculus. Differential calculus deals with infinitely small changes in some varying quantity. Let $y=f(x)$ be a function with a dependent variable $y$ and an independent variable $x$. Let $dy$ and $dx$ be the differentials. The differential forms the main part of the change in a function $y=f(x)$ as the independent variable changes. The relation between $dx$ and $dy$ is given by $dy=f^{\prime }(x)dx$.

More generally, differential calculus is used to investigate the instantaneous rate of change, for instance, velocity, to estimate the value of a small variation in a quantity, and to determine whether a function in a graph is increasing or decreasing.

Expert Answer

(a) The given function is:

$y=\tan (\sqrt{7t})$

or $y=\tan (7t{)}^{1/2}$

Here, $y$ is dependent and $t$ is an independent variable.

Taking differential of both sides using the chain rule as:

$dy={\sec }^2(7t{)}^{1/2}\cdot {\displaystyle \frac{1}{2}}(7t{)}^{-1/2}(7)dt$

Or $dy={\displaystyle \frac{7{\sec }^2(\sqrt{7t})}{2\sqrt{7t}}}dt$

(b) The given function is:

$y={\displaystyle \frac{3-v^2}{3+v^2}}$

Here, $y$ is dependent and $v$ is an independent variable.

Taking differential of both sides using the quotient rule as:

$dy={\displaystyle \frac{(3+v^2)\cdot (-2v)-(3-v^2)(2v)}{(3+v^2{)}^2}}dv$

$dy={\displaystyle \frac{-6v-v^3-6v+2v^3}{(3+v^2{)}^2}}dv$

$dy={\displaystyle \frac{-12v}{(3+v^2{)}^2}}dv$

Graph of $y={\displaystyle \frac{3-v^2}{3+v^2}}$ and its differential

Examples

Find the differential of the following functions:

(a) $f(y)=y^2-\sec (y)$

Using the power rule on first term and the chain rule on second term as:

$df(y)=[2y-\sec (y)\tan (y)]dy$

(b) $y=x^4-9x^2+12x$

Using power rule on all the terms as:

$dy=(4x^3-18x+12)dx$

(c) $h(x)=(x-2)(x-x^3)$

Rewrite the function as:

$h(x)=x^2-x^4-2x+2x^3$

$h(x)=-x^4+2x^3+x^2-2x$

Now use the power rule on all the terms as:

$dh(x)=(-4x^3+6x^2+2x-2)dx$

(d) $x={\displaystyle \frac{3}{\sqrt{t^3}}}+{\displaystyle \frac{1}{4t^4}}-{\displaystyle \frac{1}{t^{11}}}$

Rewrite the given function as:

$x=3t^{-3/2}+{\displaystyle \frac{1}{4}}{t}^{-4}-t^{-11}$

Now use power rule on all the terms as:

$dx=(-{\displaystyle \frac{9}{2}}{t}^{-1/2}-t^{-3}+11t^{-10})dt$

$dx=(-{\displaystyle \frac{9}{2\sqrt{t}}}-{\displaystyle \frac{1}{t^3}}+{\displaystyle \frac{11}{t^{10}}})dt$

(e) $y=\ln (\sin (2x))$

Using the chain rule as:

$dy={\displaystyle \frac{1}{\sin (2x)}}\cdot \cos (2x)\cdot 2dx$

$dy={\displaystyle \frac{2\cos (2x)}{\sin (2x)}}dx$

Or $dy=2\cot (2x)dx$

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