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