Question

Find the normal to the curve \[y = {\left( {1 + 2x} \right)^x} + {\sin ^{ - 1}}\left( {{{\sin }^3}\left( x \right)} \right)\] at \[x = 0\] .
A. \[y = 0\]
B. \[x + y = 1\]
C. \[x = 0\]
D. None of these

Answer 1

Hint: First, we differentiating the given equation with different type of property to get simplify form. After that we substitute the given value \[x = 0\] and we get the value of differentiation also substitute this value \[x = 0\] in given equation to get the value of \[y\] . Then using the general form of normal of a curve we got the required solution.

Formula Used:
Differentiation property \[\dfrac{d}{{dx}}\left( {u(x) + v(x)} \right) = \dfrac{d}{{dx}}\left( {u(x)} \right) + \dfrac{d}{{dx}}\left( {v(x)} \right)\] , where \[u\left( x \right),v\left( x \right)\] are functions of \[x\] .
Differentiation formula of \[{a^x}\] , \[\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}\log a\] and
Differentiation formula of inverse, \[\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}\]
Differentiation formula of sine, \[\dfrac{d}{{dx}}\left( {{{\sin }^n}x} \right) = n{\sin ^{n - 1}}x \times \cos x\]

Complete step by step solution:
Given \[y = {\left( {1 + 2x} \right)^x} + {\sin ^{ - 1}}\left( {{{\sin }^3}\left( x \right)} \right)\] ………..(1)
Differentiating the given equation with respect to \[x\] and simplifying, we get
\[\dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {{{\left( {1 + 2x} \right)}^x} + {{\sin }^{ - 1}}\left( {{{\sin }^3}\left( x \right)} \right)} \right)\]……….(2)
Using differentiation property, \[\dfrac{d}{{dx}}\left( {u(x) + v(x)} \right) = \dfrac{d}{{dx}}\left( {u(x)} \right) + \dfrac{d}{{dx}}\left( {v(x)} \right)\] in above equation (2) and we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {{{\left( {1 + 2x} \right)}^x}} \right) + \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}\left( {{{\sin }^3}\left( x \right)} \right)} \right)\] ………………(3)
Now we use the differentiating formula \[\dfrac{d}{{dx}}\left( {{a^x}} \right) = {a^x}\log a\] in above equation (3) and we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( y \right) = {\left( {1 + 2x} \right)^x}\log \left( {1 + 2x} \right) + \dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}\left( {{{\sin }^3}\left( x \right)} \right)} \right)\] ……………(4)
Using the differentiating formula \[\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}x} \right) = \dfrac{1}{{\sqrt {1 - {x^2}} }}\] in above equation (4) and we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( y \right) = {\left( {1 + 2x} \right)^x}\log \left( {1 + 2x} \right) + \dfrac{1}{{\sqrt {1 - {{\sin }^3}\left( x \right)} }}\dfrac{d}{{dx}}\left( {{{\sin }^3}\left( x \right)} \right)\] …………..(5)
Using the formula \[\dfrac{d}{{dx}}\left( {{{\sin }^n}x} \right) = n{\sin ^{n - 1}}x \times \cos x\] in above equation (5) and we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( y \right) = {\left( {1 + 2x} \right)^x}\log \left( {1 + 2x} \right) + \dfrac{1}{{\sqrt {1 - {{\sin }^3}\left( x \right)} }}\left( {3{{\sin }^2}x \times \cos x} \right)\] …………..(6)
Now given the value of \[x = 0\]
Substitute the value of \[x\] in the above equation (6) and we get
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = {\left( {1 + 2 \times 0} \right)^0}\log \left( {1 + 2 \times 0} \right) + \dfrac{{3{{\sin }^2}0 \times \cos 0}}{{\sqrt {1 - {{\sin }^3}\left( 0 \right)} }}\]
We know the value of \[\sin 0 = 0\] and \[\cos 0 = 1\] .
Substitute \[\sin 0 = 0\] , \[\cos 0 = 1\] and we get
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = {\left( 1 \right)^0}\log \left( 1 \right) + \dfrac{{3 \times {0^2} \times 1}}{{\sqrt {1 - {0^3}} }}\]
Substitute the value \[\log 1 = 0\] and we get
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = 0 + 0\]
\[ \Rightarrow {\left( {\dfrac{{dy}}{{dx}}} \right)_{x = 0}} = 0\]
Now substitute the value \[x = 0\] in equation (1) and we get
\[ \Rightarrow y = {\left( {1 + 2 \times 0} \right)^0} + {\sin ^{ - 1}}\left( {{{\sin }^3}\left( 0 \right)} \right)\]
\[ \Rightarrow y = {\left( 1 \right)^0} + {\sin ^{ - 1}}\left( 0 \right)\]
\[ \Rightarrow y = 1 + 0\]
\[ \Rightarrow y = 1\]
The general form of the equation of normal of a line
\[\dfrac{{y - {y_0}}}{{x - {x_0}}} = - \dfrac{{dy}}{{dx}}\] , where \[\left( {{x_0},{y_0}} \right) = \left( {0,1} \right)\]
\[ \Rightarrow \dfrac{{x - 0}}{{y - 1}} = - 0\]
Cross multiplying and we get
\[ \Rightarrow x = 0\]
Therefore, the normal of the curve \[y = {\left( {1 + 2x} \right)^x} + {\sin ^{ - 1}}\left( {{{\sin }^3}\left( x \right)} \right)\] is \[x = 0\] .

Hence the correct option is C.

Note: Students got confused while they using the form of normal equation \[y - {y_0} = - \dfrac{1}{{\dfrac{{dy}}{{dx}}}}\left( {x - {x_0}} \right)\] because \[\dfrac{1}{{\dfrac{{dy}}{{dx}}}}\] gives us error value. We need to reframe this formula and use it to get the equation of normal.