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Question: Find the derivative of \(\sqrt {ax + b} \) with respect to \(x\), using the first principles of diff...

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

Explanation

Solution

Given an expression and we have to differentiate the expression concerning the variable xx. We have to differentiate the expression using the first principle. First, we will find the general expression for the slope for any value of x. This slope is written as change in y divided by change in x. Then, the derivative from first principles gives the instantaneous rate of change of quantity y with respect to variable x.

Formula used:
The slope of the tangent line is given as:
dydx=lim⁡h→0f(x+h)−f(x)h\dfrac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {x + h} \right) - f\left( x \right)}}{h}

Complete step-by-step answer:
We are given the expressionax+b\sqrt {ax + b} . First, we will find the value of f(x+h)f\left( {x + h} \right) by substituting x+hx + h for xx into the expression.
f(x+h)=a(x+h)+bf\left( {x + h} \right) = \sqrt {a\left( {x + h} \right) + b}
Now we will find the slope of the tangent line by substituting the values into the formula.
f′(x)=lim⁡h→0a(x+h)+b−ax+bhf'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {a\left( {x + h} \right) + b} - \sqrt {ax + b} }}{h}
On directly substituting the value h=0h = 0, the expression gives divide by zero error.
Thus, we will rationalize the numerator by multiplying and dividing the expression by conjugate of the numerator.
f′(x)=lim⁡h→0a(x+h)+b−ax+bh×a(x+h)+b+ax+ba(x+h)+b+ax+bf'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {a\left( {x + h} \right) + b} - \sqrt {ax + b} }}{h} \times \dfrac{{\sqrt {a\left( {x + h} \right) + b} + \sqrt {ax + b} }}{{\sqrt {a\left( {x + h} \right) + b} + \sqrt {ax + b} }}
Apply the algebraic identity (a−b)(a+b)=a2−b2\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2} to the numerator of the expression, we get:
f′(x)=lim⁡h→0(a(x+h)+b)2−(ax+b)2h(a(x+h)+b+ax+b)f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \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)}}
On further simplifying the expression, we get:
f′(x)=lim⁡h→0a(x+h)+b−ax−bh(a(x+h)+b+ax+b)f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{a\left( {x + h} \right) + b - ax - b}}{{h\left( {\sqrt {a\left( {x + h} \right) + b} + \sqrt {ax + b} } \right)}}
f′(x)=lim⁡h→0ax+ah+b−ax−bh(a(x+h)+b+ax+b)f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{ax + ah + b - ax - b}}{{h\left( {\sqrt {a\left( {x + h} \right) + b} + \sqrt {ax + b} } \right)}}
Now, we will cancel out the common terms in the expression.
f′(x)=lim⁡h→0ax+ah+b̸−ax−b̸h(a(x+h)+b+ax+b)f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{{ax}} + ah + \not b - {{ax}} - \not b}}{{h\left( {\sqrt {a\left( {x + h} \right) + b} + \sqrt {ax + b} } \right)}}
f′(x)=lim⁡h→0ahh(a(x+h)+b+ax+b)f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{{a{h}}}{{{h}\left( {\sqrt {a\left( {x + h} \right) + b} + \sqrt {ax + b} } \right)}}
f′(x)=lim⁡h→0aa(x+h)+b+ax+bf'\left( x \right) = \mathop {\lim }\limits_{h \to 0} \dfrac{a}{{\sqrt {a\left( {x + h} \right) + b} + \sqrt {ax + b} }}
Now, we will substitute the limit h=0h = 0 to the expression:
f′(x)=aa(x+0)+b+ax+bf'\left( x \right) = \dfrac{a}{{\sqrt {a\left( {x + 0} \right) + b} + \sqrt {ax + b} }}
On further simplifying the expression, we get:
f′(x)=aax+b+ax+bf'\left( x \right) = \dfrac{a}{{\sqrt {ax + b} + \sqrt {ax + b} }}
f′(x)=a2ax+bf'\left( x \right) = \dfrac{a}{{2\sqrt {ax + b} }}

Final answer: Hence, the derivative of the expression ax+b\sqrt {ax + b} is a2ax+b\dfrac{a}{{2\sqrt {ax + b} }}

Note:
In such types of questions the students mainly don't get an approach on how to solve it. In such types of questions students mainly forget to apply the rationalization of the expression if the expression gives divide by zero error on directly substituting the limits to the expression.