Question

How do you find the derivative of y = $\ln (70{{x}^{2}}+24x+7)$?

Answer 1

To solve the above stated question we will have to use the chain rule property as we can see that y consists of two concentric functions i.e. f(x) = g(h(x)) and to find the derivative of this function easily chain rule is an easy method to be applied.

Complete step by step solution:
Let y = f(x)
Now with the help of chain rule we can solve the above question.
Chain rule states that:
If F(x) = f(g(x))
Then F’(x) = f’ (g(x)) g’(x)
So to use the chain rule property In our question we will be using it in the format that has been stated below:
f(x) = g(h(x))
⇒f’(x) = g’ (h(x)) h’(x)………(1)
In the above stated functions the values of those functions are stated below which will then further be used in the upcoming calculations:
g (h(x)) = $\ln (70{{x}^{2}}+24x+7)$
⇒h(x) = $70{{x}^{2}}+24x+7$
⇒f’(x) = $\dfrac{df}{dx}$ (this defines the derivative of f(x))
⇒g’(h(x)) = $\dfrac{dg}{dx}$ (this defines the derivative of g(h(x)))
⇒h’(x) = $\dfrac{dh}{dx}$ (this defines the derivative of h(x))
As we know that the differentiation of ln(x) = $\dfrac{1}{x}$
By using the above stated property (or formula) we will find the value of g’(h(x))
So after using this property we got,
⇒g’(h(x)) = $\dfrac{1}{70{{x}^{2}}+24x+7}$
Now, we have to find the value of h’(x), so, we will be using the sum rule of derivation which is stated as below:
If F(x) = g(x) + h(x) + t(x) + and so on
Then F’(x) = g’(x) + h’(x) + t’(x) + and so on
As we know that the differentiation of ${{x}^{2}}=\text{ }2x$
Differentiation of x = 1
Differentiation of constant = 0
By using all the above stated property (or formula) we will find the value of h’(x)
So after using this property we got,
⇒h’(x) = 70(2x) + 24(1) + 7(0)
⇒h’(x) = 140x +24
Now we will be substituting the values of g’ (h(x)) and h’(x) which we got from solving in equation (1) from which we will get the derivative of the function which was initially asked in the question as:
⇒f’(x) = $\dfrac{140x+24}{70{{x}^{2}}+24x+7}$
So, from the above equation we can clearly say that the derivative of y = $\ln (70{{x}^{2}}+24x+7)$is $\dfrac{140x+24}{70{{x}^{2}}+24x+7}$

Note:
The point where mistakes are generally made is to how the chain rule should be applied. For applying chain rule we should use the rule in the descending format i.e. if f(x) = g(h(t(c(x)))) then by using chain rule property we will get f’(x) = g’(h(t(c(x)))) h’(t(c(x))) t’(c(x)) c’(x) which is the derivative of the function that was stated.