Question

How do you find the derivative of $\dfrac{4}{x+3}$

Answer 1

Now to solve this problem we will use a method of substitution. We will substitute x + 3 = t. and rewrite the expression. Now we know that differentiation of ${{x}^{n}}$ is given by $n{{x}^{n-1}}$ . We will use this to differentiate the obtained equation and then re-substitute the value of t. Hence we have the derivative of $\dfrac{4}{x+3}$ .

Complete step-by-step solution:
Now we want to find the derivative of a given expression $\dfrac{4}{x+3}$ .
Now let us say y = $\dfrac{4}{x+3}$ .
Let us substitute x + 3 = t.
Then we have $y=\dfrac{4}{t}$
Now we can write the above equation as $y=4{{t}^{-1}}$
Now we know that $\dfrac{d\left( cx \right)}{dx}=c.\dfrac{d\left( y \right)}{dx}$ where c is a constant.
Since here 4 is a constant we can take it out of differentiation
Now we know that let us differentiate the given function with respect to t
$\Rightarrow \dfrac{dy}{dt}=4\dfrac{d\left( {{t}^{-1}} \right)}{dt}$
Now we know that differentiation of ${{x}^{n}}$ is given by $n{{x}^{n-1}}$ for any integer n.
Hence we get,
$\begin{aligned} & \Rightarrow \dfrac{dy}{dt}=4.\left( -1 \right){{t}^{-1-1}} \\\ & \Rightarrow \dfrac{dy}{dt}=-4{{t}^{-2}} \\\ \end{aligned}$
Now let us re-substitute the value of t as x + 3 Hence, we get $\dfrac{dy}{dx}=-4{{\left( x+3 \right)}^{-2}}$
Now we can rewrite the above equation as
$\dfrac{dy}{dx}=\dfrac{-4}{{{\left( x+3 \right)}^{2}}}$
Now we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ .
Hence, using this we get
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-4}{{{x}^{2}}+6x+9}$
Hence the differentiation of $\dfrac{4}{x+3}$ is $\dfrac{-4}{{{x}^{2}}+6x+9}$.

Note: Now for the given problem we can also evaluate it by using definition of limits. By definition of limits we have $f'\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ . Now here we will substitute the function and simplify the expression by taking LCM and solving the limits and hence get the derivative of the given function. Also note that here we have x + 3 = t. differentiating on both sides we get dx = dt. Hence we have used $\dfrac{dy}{dt}=\dfrac{dy}{dx}$ above.