Question

The slope of normal to the curve \[y = \log \left( {\log {\rm{ x}}} \right)\] at \[x = e\] is
\[\begin{array}{l}A.\,\,e\\B.\,\, - e\\C.\,\,\dfrac{1}{e}\\D.\,\, - \dfrac{1}{e}\end{array}\]

Answer 1

Hint: In the question, at first, find the slope by differentiating the given function using chain rule which is if \[f\left( x \right) = j\left( {k\left( x \right)} \right)\] then, on differentiating it, we get, \[f'\left( x \right) = j'\left( {k\left( x \right)} \right) \times k'\left( x \right)\], where \[j'\left( {k\left( x \right)} \right)\] is differentiation of j(x) keeping k(x) constant and \[k'\left( x \right)\] is differentiation of k(x) irrespective of what j(x) is. Then, divide it by (-1) to get the slope of normal.

Complete step-by-step answer:
In the question, we are given a function of a curve which is \[y = \log \left( {\log {\rm{ x}}} \right)\] and we have to find normal of the curve at a given x-point which is e.
So, before finding it let us know how to find slope, then, we will divide it from -1.
Now, let’s suppose a function f(x) is given, then, we can find its slope at any point by differentiating it with respect to x and then for finding slope at a particular point x=x, we will substitute it to get the answer. After finding the slope at a given point, we can find the normal at the same point by dividing it from -1.
The function in the given question is, log (log x), let it be written as,
\[f\left( x \right) = \log \left( {\log {\rm{ x}}} \right)\]
Which can be written as,
\[f\left( x \right) = g\left( {g\left( x \right)} \right)\]
Where, g(x) is log x,
So, we can differentiate f(x) using chain rule if,
\[f\left( x \right) = j\left( {k\left( x \right)} \right)\]
We can write the derivative as
\[f'\left( x \right) = j'\left( {k\left( x \right)} \right) \times k'\left( x \right)\]
Hence, here \[j'\left( {k\left( x \right)} \right)\] represents differentiation of j(x) keeping k(x) constant and \[k'\left( x \right)\] is differentiation of k(x) irrespective of what k(x) is.
So, here \[f\left( x \right) = g\left( {g\left( x \right)} \right)\] then, according to chain rule,
\[f'\left( x \right) = g'\left( {g\left( x \right)} \right) \times g'\left( x \right)\]
Here, \[g\left( x \right) = \log {\rm{ x, so, g'}}\left( x \right) = \dfrac{1}{x}{\rm{ and g'}}\left( {g\left( x \right)} \right) = \dfrac{1}{{\log \left( x \right)}}\]
So, we can say that,
\[\begin{array}{l}f'\left( x \right) = \dfrac{1}{{\log \left( {\rm{x}} \right)}} \times \dfrac{1}{x}\\ \Rightarrow f'\left( x \right) = \dfrac{1}{{x\log {\rm{x}}}}\end{array}\]
Here, the point is given x=e, so, the value of \[f'\left( e \right) = \dfrac{1}{{e\log e}} \Rightarrow \dfrac{1}{e}\]
Here, slope is \[\dfrac{1}{e}\] so, the value of normal will be \[\dfrac{{\left( { - 1} \right)}}{{\left( {\dfrac{1}{e}} \right)}} \Rightarrow - e\]
Hence, the correct option is B.

Note: Generally students while solving the slope of normal, they miss out and leave the question after finding the slope of the tangent, thus, they should read the question twice before attempting it.