Question

What is the derivative of $f(x)=\ln \left( {{x}^{2}}+3x-5 \right)$.

Answer 1

For solving this question we have to know the proper application of chain rule and some basic derivative formulas. We have to know that $\dfrac{d}{dx}\ln a=\dfrac{1}{a}\times \dfrac{d}{dx}a$. where $a$ is a variable. Also, we should avoid calculation mistakes to get accurate answers.

Complete step by step solution:
From the question we were given that
\Rightarrow $$$$f(x)=\ln \left( {{x}^{2}}+3x-5 \right)…………..(1)
From the question it is clear that we have find derivative of $\ln \left( {{x}^{2}}+3x-5 \right)$
Now let us assume $f\left( x \right)=y$
So the equation (1) becomes $y=\ln \left( {{x}^{2}}+3x-5 \right)$……………(2)
So, now we have to find $\dfrac{dy}{dx}$.
Take the equation $y=\ln \left( {{x}^{2}}+3x-5 \right)$
Now let us differentiate on both sides,
So, we will get equation as
\Rightarrow $$$$\dfrac{d}{dx}y=\dfrac{d}{dx}\ln \left( {{x}^{2}}+3x-5 \right)……………..(3)
From the basic derivative formulas, we know that $\dfrac{d}{dx}\ln a=\dfrac{1}{a}\times \dfrac{d}{dx}a$. where $a$is variable
So now let us try to apply this formula for solving equation (3)
So, We will get
\Rightarrow $$$$\dfrac{d}{dx}\ln \left( {{x}^{2}}+3x-5 \right)=\dfrac{1}{{{x}^{2}}+3x-5}\times \dfrac{d}{dx}\left( {{x}^{2}}+3x-5 \right)………….(4)
Now we will apply chain rule to solve. $\dfrac{d}{dx}\left( {{x}^{2}}+3x-5 \right)$
From the chain rule we know that
\Rightarrow $$$$\dfrac{d}{dx}\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}}+............+{{a}_{n}} \right)=\dfrac{d}{dx}{{a}_{1}}+\dfrac{d}{dx}{{a}_{2}}+\dfrac{d}{dx}{{a}_{3}}+............+\dfrac{d}{dx}{{a}_{n}}
Now we can write
$\dfrac{d}{dx}\left( {{x}^{2}}+3x-5 \right)=\dfrac{d}{dx}{{x}^{2}}+\dfrac{d}{dx}3x-\dfrac{d}{dx}5$…………….(5)
We know that derivative of constant is ZERO, so $\dfrac{d}{dx}5=0$.
We know that $\dfrac{d}{dx}{{a}^{2}}=2a\times \dfrac{da}{dx}$. where $a$is variable
So, we can write $\dfrac{d}{dx}{{x}^{2}}=2x\times \dfrac{dx}{dx}$ .as $\dfrac{dx}{dx}=1$ it can be modified as $\dfrac{d}{dx}{{x}^{2}}=2x$
so, $\dfrac{d}{dx}{{x}^{2}}=2x$.
Also, we know that $\dfrac{d}{dx}ka=k\times \dfrac{d}{dx}a$,where $a$is variable and $k$=constant
So, we can write $\dfrac{d}{dx}3x=3\times 1$ as $\left[ \dfrac{dx}{dx}=1 \right]$
So (5) can be written as $\dfrac{d}{dx}\left( {{x}^{2}}+3x-5 \right)=2x+3-0$
\Rightarrow $$$$\dfrac{d}{dx}\left( {{x}^{2}}+3x-5 \right)=2x+3……………………(6)
Put (6) in (4) we get
$\Rightarrow$ $\dfrac{d}{dx}\ln \left( {{x}^{2}}+3x-5 \right)=\dfrac{1}{{{x}^{2}}+3x-5}\times \left( 2x+3 \right)$
\Rightarrow $$$$\dfrac{d}{dx}\ln \left( {{x}^{2}}+3x-5 \right)=\dfrac{2x+3}{{{x}^{2}}+3x-5}
So, finally we can conclude that the derivative of $\ln \left( {{x}^{2}}+3x-5 \right)$ is $\dfrac{2x+3}{{{x}^{2}}+3x-5}$
\Rightarrow $$$$\dfrac{d}{dx}f(x)= $\dfrac{2x+3}{{{x}^{2}}+3x-5}$

Note: While solving this type of questions students should use correct formulas. using wrong formulas can lead to wrong answers. Students may have misconception that $\dfrac{d}{dx}\ln a=a\left( \ln a-1 \right)\dfrac{d}{dx}a$, which is formula for integration of $\ln x$ which is completely WRONG concept and the RIGHT formula is $\dfrac{d}{dx}\ln a=\dfrac{1}{a}\times \dfrac{d}{dx}a$