
If \[y = \log [\sin ({x^2})],0 < x < \dfrac{\pi }{2}\], then what is the value of \[\dfrac{{dy}}{{dx}}\] at \[x = \dfrac{{\sqrt \pi }}{2}\]?
A. 0
B. 1
C. \[\dfrac{\pi }{4}\]
D. \[\sqrt \pi \]
Answer
164.1k+ views
Hint: In solving the above question, first we will find the derivative of the function using derivative formulas and chain rule, then substitute the value of \[x\] given and after simplifying the result we will get the desired value.
Formula used :
We will use derivative formulas \[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\], \[\dfrac{d}{{dx}}\sin x = \cos x\], \[\dfrac{d}{{dx}}{x^n} = n \cdot {x^{n - 1}}\] and we will be using chain rule, trigonometric identity \[\dfrac{{\cos x}}{{\sin x}} = \cot x\], and \[\cot \dfrac{\pi }{4} = 1\].
Complete Step-by- Step Solution:
Given \[y = \log [\sin ({x^2})],0 < x < \dfrac{\pi }{2}\]
Now we will apply differentiation on both sides, we will get,
\[ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\log [\sin ({x^2})]\]
Now we will apply derivative formulas, and we will use the chain rule, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin {x^2}}}\dfrac{d}{{dx}}\sin {x^2}\]
Now we will again apply derivative formulas, and we will use the chain rule, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin {x^2}}}\left( {\cos {x^2}} \right)\dfrac{d}{{dx}}{x^2}\]
Now we will again apply derivative formulas, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin {x^2}}}\left( {\cos {x^2}} \right)\left( {2x} \right)\]
Now using trigonometric identity i.e.,\[\dfrac{{\cos x}}{{\sin x}} = \cot x\], we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 2x\cot {x^2}\]
Now we will substitute the value of \[x\] given in the expression i.e., \[x = \dfrac{{\sqrt \pi }}{2}\], we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 2\dfrac{{\sqrt \pi }}{2}\cot {\left( {\dfrac{{\sqrt \pi }}{2}} \right)^2}\]
Now we will simplify the expression we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \sqrt \pi \cot \left( {\dfrac{\pi }{4}} \right)\]
Now using trigonometric table for \[\cot \left( {\dfrac{\pi }{4}} \right)\] value we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \sqrt \pi \left( 1 \right)\]
Now we will further simplify we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \sqrt \pi \]
The correct option is D.
Note: Students got confused when they differentiate some functions of functions. e.g., \[y = u\left( {v\left( x \right)} \right)\] . This type of function can differentiate with the chain rule.
Chain rule:
If a function given \[y = u\left( {v\left( x \right)} \right)\]
Then the required expression after differentiate look like
\[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dv}} \times \dfrac{{dv}}{{dx}}\]
Using the chain rule we need to simplify the required answer.
Formula used :
We will use derivative formulas \[\dfrac{d}{{dx}}\log x = \dfrac{1}{x}\], \[\dfrac{d}{{dx}}\sin x = \cos x\], \[\dfrac{d}{{dx}}{x^n} = n \cdot {x^{n - 1}}\] and we will be using chain rule, trigonometric identity \[\dfrac{{\cos x}}{{\sin x}} = \cot x\], and \[\cot \dfrac{\pi }{4} = 1\].
Complete Step-by- Step Solution:
Given \[y = \log [\sin ({x^2})],0 < x < \dfrac{\pi }{2}\]
Now we will apply differentiation on both sides, we will get,
\[ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\log [\sin ({x^2})]\]
Now we will apply derivative formulas, and we will use the chain rule, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin {x^2}}}\dfrac{d}{{dx}}\sin {x^2}\]
Now we will again apply derivative formulas, and we will use the chain rule, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin {x^2}}}\left( {\cos {x^2}} \right)\dfrac{d}{{dx}}{x^2}\]
Now we will again apply derivative formulas, we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sin {x^2}}}\left( {\cos {x^2}} \right)\left( {2x} \right)\]
Now using trigonometric identity i.e.,\[\dfrac{{\cos x}}{{\sin x}} = \cot x\], we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 2x\cot {x^2}\]
Now we will substitute the value of \[x\] given in the expression i.e., \[x = \dfrac{{\sqrt \pi }}{2}\], we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 2\dfrac{{\sqrt \pi }}{2}\cot {\left( {\dfrac{{\sqrt \pi }}{2}} \right)^2}\]
Now we will simplify the expression we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \sqrt \pi \cot \left( {\dfrac{\pi }{4}} \right)\]
Now using trigonometric table for \[\cot \left( {\dfrac{\pi }{4}} \right)\] value we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \sqrt \pi \left( 1 \right)\]
Now we will further simplify we will get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \sqrt \pi \]
The correct option is D.
Note: Students got confused when they differentiate some functions of functions. e.g., \[y = u\left( {v\left( x \right)} \right)\] . This type of function can differentiate with the chain rule.
Chain rule:
If a function given \[y = u\left( {v\left( x \right)} \right)\]
Then the required expression after differentiate look like
\[\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dv}} \times \dfrac{{dv}}{{dx}}\]
Using the chain rule we need to simplify the required answer.
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