Question

How do you find the derivative of the function using the definition of derivative $g(t)=\dfrac{7}{\sqrt{t}}$?

Answer 1

Hint: In this question we will first write the definition of derivative and the expression of how to calculate the derivative of a function, as a function of limit. We will then substitute the value of $g(t)$ from the given question in the expression and simplify the terms. We will then rationalize the numerator to substitute the value of limit and get the required solution.

Complete step by step solution:
We have the given function as:
$g(t)=\dfrac{7}{\sqrt{t}}$
Now the general definition of derivative as a function of limit is given as:
$g'(t)=\displaystyle \lim_{h \to 0}\dfrac{g(t+h)-g(t)}{h}$
Now to find the derivative, we will substitute the value of $g(t)$ in the expression. On substituting, we get:
$g'(t)=\displaystyle \lim_{h \to 0}\dfrac{\dfrac{7}{\sqrt{t+h}}-\dfrac{7}{\sqrt{t}}}{h}$
Now the numerator has two fractions. On taking the lowest common multiple in the numerator by cross multiplying, we get:
$g'(t)=\displaystyle \lim_{h \to 0}\dfrac{\dfrac{7\sqrt{t}-7\sqrt{t+h}}{\sqrt{t}\sqrt{t+h}}}{h}$
On rearranging the terms in the expression, we get:
$g'(t)=\displaystyle \lim_{h \to 0}\dfrac{7\sqrt{t}-7\sqrt{t+h}}{h\sqrt{t}\sqrt{t+h}}$
On taking out $7$ common in the numerator, we get:
$g'(t)=\displaystyle \lim_{h \to 0}\dfrac{7\left( \sqrt{t}-\sqrt{t+h} \right)}{h\sqrt{t}\sqrt{t+h}}$
Now it is to be noted that when we solve the limit by substituting the value of $h=0$ in the expression, we get the expression in the form of $\dfrac{0}{0}$ which is the indeterminate form therefore, we will rationalize the numerator $\sqrt{t}-\sqrt{t+h}$ by multiplying and dividing by its conjugate which is $\sqrt{t}+\sqrt{t+h}$.
On multiplying, we get:
$g'(t)=\displaystyle \lim_{h \to 0}\dfrac{7\left( \sqrt{t}-\sqrt{t+h} \right)}{h\sqrt{t}\sqrt{t+h}}\times \dfrac{\sqrt{t}+\sqrt{t+h}}{\sqrt{t}+\sqrt{t+h}}$
On multiplying the terms in the numerator, we get:
$g'(t)=\displaystyle \lim_{h \to 0}\dfrac{7\left( t-\left( t+h \right) \right)}{h\sqrt{t}\sqrt{t+h}\left( \sqrt{t}+\sqrt{t+h} \right)}$
On simplifying, we get:
$g'(t)=\displaystyle \lim_{h \to 0}\dfrac{-7h}{h\sqrt{t}\sqrt{t+h}\left( \sqrt{t}+\sqrt{t+h} \right)}$
On cancelling $h$, we get:
$g'(t)=\displaystyle \lim_{h \to 0}\dfrac{-7}{\sqrt{t}\sqrt{t+h}\left( \sqrt{t}+\sqrt{t+h} \right)}$
Since $h$ has been eliminated in the multiplication form, we won’t get the limit as an indeterminate value therefore, on substituting $h=0$ we get:
$g'(t)=\dfrac{-7}{\sqrt{t}\sqrt{t+0}\left( \sqrt{t}+\sqrt{t+0} \right)}$
On simplifying, we get:
$g'(t)=\dfrac{-7}{t\left( \sqrt{t}+\sqrt{t} \right)}$
Which can be written as:
$g'(t)=\dfrac{-7}{2t\sqrt{t}}$, which is the required solution.

Note: It is to be remembered that the general formula for derivative of $\sqrt{x}$ can be written as $\dfrac{d}{dx}\sqrt{x}=\dfrac{1}{2\sqrt{x}}$ and that the derivative of $\dfrac{1}{\sqrt{x}}$ can be written as $\dfrac{d}{dx}\dfrac{1}{\sqrt{x}}=\dfrac{-1}{2{{x}^{3}}}$. It is to be remembered that whenever a constant is multiplied in a derivative, it is to be taken out.