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Question: Find the \(\dfrac{dy}{dx}\) for the following equations: (a) \(y=1+3x-{{x}^{2}}\) (b) \(y={{\le...

Find the dydx\dfrac{dy}{dx} for the following equations:
(a) y=1+3x−x2y=1+3x-{{x}^{2}}
(b) y=(ax+b)2y={{\left( ax+b \right)}^{2}}

Explanation

Solution

We explain the concept of derivation of a dependent variable with respect to an independent variable. We first find the formula for the derivation for nth{{n}^{th}} power of a variable x where ddx(xn)=nxn−1\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}. We place the value for nn. We get solutions for the derivatives of y=1+3x−x2y=1+3x-{{x}^{2}} and y=(ax+b)2y={{\left( ax+b \right)}^{2}}. We also explain the theorem with the help of the first order derivative.

Complete step by step answer:
Differentiation, the fundamental operations in calculus, deals with the rate at which the dependent variable changes with respect to the independent variable. The measurement quantity of its rate of change is known as derivative or differential coefficients. We find the increment of those variables for small changes. We mathematically express it as dydx\dfrac{dy}{dx} where y=f(x)y=f\left( x \right).

(a) The formula of derivation for nth{{n}^{th}} power of a variable x is ddx(xn)=nxn−1\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}. The condition being n\in \mathbb{R}\backslash \left\\{ 0 \right\\}. For our given function y=1+3x−x2y=1+3x-{{x}^{2}}, we use values of n as 0, 1, 2. We apply the theorem and get
dydx=ddx(1+3x−x2) ⇒dydx=0+3×1−2x ∴dydx=3−2x\dfrac{dy}{dx}=\dfrac{d}{dx}\left( 1+3x-{{x}^{2}} \right) \\\ \Rightarrow \dfrac{dy}{dx}=0+3\times 1-2x \\\ \therefore \dfrac{dy}{dx} =3-2x

(b) Now we find the derivative of y=(ax+b)2y={{\left( ax+b \right)}^{2}} using the chain rule.So,

\Rightarrow \dfrac{dy}{dx}=2\left( ax+b \right)\dfrac{d\left( ax+b \right)}{dx} \\\ \therefore \dfrac{dy}{dx} =2a\left( ax+b \right)$$ **Therefore, the derivative of the functions $y=1+3x-{{x}^{2}}$ and $y={{\left( ax+b \right)}^{2}}$ is $3-2x$ and $$2a\left( ax+b \right)$$ respectively.** **Note:** If the ratio of $\dfrac{\Delta y}{\Delta x}$ tends to a definite finite limit when $$\Delta x\to 0$$, then the limiting value obtained by this can also be found by first order derivative. We can also apply a first order derivative theorem to get the differentiated value of ${{x}^{n}}$. We know that $\dfrac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$. Here $f\left( x \right)={{x}^{n}}$. Also, $f\left( x+h \right)={{\left( x+h \right)}^{n}}$. We assume $x+h=u$ which gives $f\left( u \right)={{\left( u \right)}^{n}}$ and $h=u-x$. As $h\to 0$ we get $u\to x$. So, $\dfrac{df}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}=\underset{u\to x}{\mathop{\lim }}\,\dfrac{{{u}^{n}}-{{x}^{n}}}{u-x}$. We know the limit value $\underset{x\to a}{\mathop{\lim }}\,\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n-1}}$. Therefore, $$\dfrac{df}{dx}=\underset{u\to x}{\mathop{\lim }}\,\dfrac{{{u}^{n}}-{{x}^{n}}}{u-x}=n{{x}^{n-1}}$$.