Question

Find the derivative of $\sqrt{ax+b}$ with respect to $x$ , using the first principles of differentiation.

Answer 1

To calculate the derivative of the function $f\left( x \right)=\sqrt{ax+b}$ using the first principle of differentiation, use the formula, $\dfrac{df\left( x \right)}{dx}={\displaystyle \lim_{h\to 0}}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ . Now, solve it further and calculate the value of the limit.

Complete step-by-step solution
According to the question, we are given a function $f\left( x \right)$ and we have to find its derivative using the first principles of differentiation.
The given function, $f\left( x \right)=\sqrt{ax+b}$ ……………………………..(1)
Here, we require the first principle of differentiation.
We know the first principles of differentiation that if there is a function $f\left( x \right)$ , then its rate of change with respect to x is, $\dfrac{df\left( x \right)}{dx}={\displaystyle \lim_{h\to 0}}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ …………………………………………(2)
Now, from equation (1) and on substituting $f\left( x \right)$ by $\sqrt{ax+b}$ in equation (2), we get
$\Rightarrow \dfrac{df\left( x \right)}{dx}={\displaystyle \lim_{h\to 0}}\dfrac{\sqrt{a\left( x+h \right)+b}-\sqrt{ax+b}}{h}$ ……………………………………….(3)
We can observe that the above equation is complex to proceed further. Therefore, we need to simplify it in an easy form so that the limit value can be calculated.
Let us multiply by the term $\sqrt{a\left( x+h \right)+b}+\sqrt{ax+b}$ in the numerator and denominator of equation (3).
On multiplying and solving it further, we get

& \Rightarrow \dfrac{df\left( x \right)}{dx}={\displaystyle \lim_{h\to 0}}\left\\{ \dfrac{\left( \sqrt{a\left( x+h \right)+b}-\sqrt{ax+b} \right)\times \left( \sqrt{a\left( x+h \right)+b}+\sqrt{ax+b} \right)}{h\left( \sqrt{a\left( x+h \right)+b}+\sqrt{ax+b} \right)} \right\\} \\\ & \Rightarrow \dfrac{df\left( x \right)}{dx}={\displaystyle \lim_{h\to 0}}\left\\{ \dfrac{{{\left( \sqrt{a\left( x+h \right)+b} \right)}^{2}}-{{\left( \sqrt{ax+b} \right)}^{2}}}{h\left( \sqrt{a\left( x+h \right)+b}+\sqrt{ax+b} \right)} \right\\} \\\ \end{aligned}$$ $$\begin{aligned} & \Rightarrow \dfrac{df\left( x \right)}{dx}={\displaystyle \lim_{h\to 0}}\left\\{ \dfrac{a\left( x+h \right)+b-\left( ax+b \right)}{h\left( \sqrt{a\left( x+h \right)}+\sqrt{ax+b} \right)} \right\\} \\\ & \Rightarrow \dfrac{df\left( x \right)}{dx}={\displaystyle \lim_{h\to 0}}\left\\{ \dfrac{ax+ah+b-ax-b}{h\left( \sqrt{a\left( x+h \right)}+\sqrt{ax+b} \right)} \right\\} \\\ & \Rightarrow \dfrac{df\left( x \right)}{dx}={\displaystyle \lim_{h\to 0}}\left\\{ \dfrac{ah}{h\left( \sqrt{a\left( x+h \right)}+\sqrt{ax+b} \right)} \right\\} \\\ \end{aligned}$$ $$\Rightarrow \dfrac{df\left( x \right)}{dx}={\displaystyle \lim_{h\to 0}}\left( \dfrac{a}{\left( \sqrt{ax+ah+b}+\sqrt{ax+b} \right)} \right)$$ ……………………………………..(4) Since $$h$$ is tending to zero so, the value of $$ah$$ must also be tending towards to zero ………………………….(5) Now, from equation (4) and equation (5), we get $$\begin{aligned} & \Rightarrow \dfrac{df\left( x \right)}{dx}={\displaystyle \lim_{h\to 0}}\left( \dfrac{a}{\left( \sqrt{ax+0+b}+\sqrt{ax+b} \right)} \right) \\\ & \Rightarrow \dfrac{df\left( x \right)}{dx}=\left\\{ \dfrac{a}{\left( \sqrt{ax+b}+\sqrt{ax+b} \right)} \right\\} \\\ & \Rightarrow \dfrac{df\left( x \right)}{dx}=\left\\{ \dfrac{a}{2\left( \sqrt{ax+b} \right)} \right\\} \\\ \end{aligned}$$ Here, we have calculated the derivative of the function $$f\left( x \right)$$ using first principles of differentiation. Therefore, using the first principle of differentiation, the derivative of the function $$f\left( x \right)=\sqrt{ax+b}$$ is $$\dfrac{a}{2\left( \sqrt{ax+b} \right)}$$ . **Note:** To solve this type of question, one should keep one point into consideration i.e., the basic definition of first principle of differentiation. According to the first principle of differentiation, if there is a function $$f\left( x \right)$$ , then its rate of change with respect to x is, $$\dfrac{df\left( x \right)}{dx}={\displaystyle \lim_{h\to 0}}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$$.