Question

How do you find the derivative of $\ln (1 + {x^4})$ ?

Answer 1

Hint : In the given question we have to differentiate $\ln (1 + {x^4})$ with respect to x, differentiation is said to be a process of dividing a whole quantity into very small ones. We will first differentiate the whole quantity $\ln (1 + {x^4})$ and then differentiate the quantity in the parenthesis as it is also a function of x $1 + {x^4}$ . The result of multiplying these two differentiations will give us the derivative of the given function. On further solving, we will get the correct answer.

Complete step-by-step answer :
We have to find the derivative of $\ln (1 + {x^4})$
Let
$y = \ln (1 + {x^4})$
Differentiating both the sides, we get –
$\dfrac{{dy}}{{dx}} = \dfrac{{d[\ln (1 + {x^4})]}}{{dx}}$
We know that $\dfrac{{d(\ln x)}}{{dx}} = \dfrac{1}{x}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {x^4}}}\dfrac{{d(1 + {x^4})}}{{dx}} \\\ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {x^4}}}(4{x^3})\,\,\,(\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}) \;$
Hence the derivative of $\ln (1 + {x^4})$ is $\dfrac{{4{x^3}}}{{1 + {x^4}}}$ .
So, the correct answer is “ $\dfrac{{4{x^3}}}{{1 + {x^4}}}$ ”.

Note : Usually, the rate of change of something is observed over a specific duration of time, but if we have to find the instantaneous rate of change of a quantity then we differentiate it, in the expression $\dfrac{{dy}}{{dx}}$ , $dy$ represents a very small change in the quantity and $dx$ represents the small change in the quantity with respect to which the given quantity is changing. In the given question, we have a function of x, so by putting different values of x, we can find the instantaneous change in x at that particular value. For finding the derivative of a function, we must know the derivatives of some basic functions like exponential functions, logarithm functions, trigonometric functions, inverse trigonometric functions, etc. Also, we have used the chain rule in this solution