Question

Let \[y\] be an implicit function of \[x\] defined by \[{{x}^{2x}}-2{{x}^{x}}\cot y-1=0\] . Then, \[{{y}^{'}}(1)\] equals:
 (a) \[1\]
(b) \[\log 2\]
(c) \[-\log 2\]
(d) \[-1\]

Answer 1

Hint:The given problem is related to differentiation of implicit function. Use the following formulae to get the required answer: \[\dfrac{d(\cot x)}{dx}=-\cos e{{c}^{2}}x\] and \[\dfrac{d({{x}^{x}})}{dx}={{x}^{x}}(1+\log x)\] .

Complete step-by-step answer:
The given expression is \[{{x}^{2x}}-2{{x}^{x}}\cot y-1=0\] .
The given expression can be rewritten as \[{{({{x}^{x}})}^{2}}-2{{x}^{x}}\cot y-1=0\] .
Clearly , we can see that it is an implicit function . Hence , we can represent it as \[f(x,y)\equiv {{({{x}^{x}})}^{2}}-2{{x}^{x}}\cot y-1=0\].
Now , we will differentiate \[f(x,y)\] with respect to \[x\] .
 \[f(x,y)\] is an implicit function. So , to differentiate it , we will apply chain rule of differentiation . According to the chain rule of differentiation , if \[R(x,y)\] is an implicit function , then the derivative of \[R(x,y)\] with respect to \[x\] is given by \[\dfrac{dR}{dx}=\dfrac{\partial R}{\partial x}+\dfrac{\partial R}{\partial y}\times \dfrac{dy}{dx}\] .
Now , we will differentiate the function with respect to \[x\] . So, on differentiating we get ,
\[2({{x}^{x}}).\dfrac{d}{dx}({{x}^{x}})-\cot y.\dfrac{\partial }{\partial x}2({{x}^{x}})-(2{{x}^{x}}\dfrac{\partial }{\partial y}(\cot y)\times \dfrac{dy}{dx})=0.........\] equation \[(1)\]
Now, we have to find the derivative of ${{x}^{x}}$ with respect to $x$ .
Let z = ${{x}^{x}}$ . Since, the function is in exponent form, let’s apply logarithms to convert it into product form. So, log z = log $\left( {{x}^{x}} \right)$ . We know, $\log {{a}^{b}}=b\log a$.
$\Rightarrow \log z=x\log x$ . Now, to differentiate with respect to x, we will use product rule of differentiation, which is given as $\dfrac{d}{dx}\left( uv \right)=uv'+u'v$ , in the right-hand side.
On differentiating with respect to x, we get:
$\dfrac{1}{z}\dfrac{dz}{dx}=x\times \dfrac{d}{dx}\left( \log x \right)+\log x$
$\Rightarrow \dfrac{1}{z}\dfrac{dz}{dx}=x\times \dfrac{1}{x}+\log x$
$\Rightarrow \dfrac{1}{z}\dfrac{dz}{dx}=1+\log x$
We substituted $z={{x}^{x}}$ .
$\Rightarrow \dfrac{d}{dx}{{x}^{x}}={{x}^{x}}\left( 1+\log x \right)$
Now , we will substitute the value of \[\dfrac{d}{dx}({{x}^{x}})\] in equation \[(1)\] .
On substituting the value of \[\dfrac{d}{dx}({{x}^{x}})\] in equation \[(1)\] , we get , \[2{{x}^{x}}[1+\log x]{{x}^{x}}-\cot y2{{x}^{x}}(1+\log x)-2{{x}^{x}}(-\cos e{{c}^{2}}y)\dfrac{dy}{dx}=0\]
\[\Rightarrow 2{{x}^{2x}}(1+\log x)-\cot y2{{x}^{x}}(1+\log x)+2{{x}^{x}}\cos e{{c}^{2}}y\dfrac{dy}{dx}=0\]
\[\Rightarrow 2{{x}^{x}}({{x}^{x}}(1+\log x)-\cot y(1+\log x))+2{{x}^{x}}\cos e{{c}^{2}}y\dfrac{dy}{dx}=0\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{2{{x}^{x}}(1+\log x)(\cot y-{{x}^{x}})}{2{{x}^{x}}\cos e{{c}^{2}}y}\]
Now , we will find the value of \[y\] at \[x=1\] .
We are given \[{{x}^{2x}}-2{{x}^{x}}\cot y-1=0.........\] equation \[(2)\] .
 To find the value of \[y\] at \[x=1\] , we will substitute \[x=1\] in equation \[(2)\] .
On substituting \[x=1\] in equation \[(2)\] , we get ,
\[{{1}^{2.1}}-{{2.1}^{1}}\cot y-1=0\]
\[\begin{align}
  & \Rightarrow -2\cot y=0 \\
 & \Rightarrow \cot y=0 \\
 & \Rightarrow y=\dfrac{\pi }{2} \\
\end{align}\]
So , at \[x=1\] , we can see the value of \[y=\dfrac{\pi }{2}\] .
Now , substituting \[x=1\] and \[y=\dfrac{\pi }{2}\] in the value of \[\dfrac{dy}{dx}\] we get ,
 \[\dfrac{dy}{dx}=\dfrac{2(1+0)(0-1)}{2\times 1}=\dfrac{-2}{2}=-1\] .
Hence , the value of the derivative of \[{{x}^{2x}}-2{{x}^{x}}\cot y-1=0\] at \[x=1\] is equal to \[-1\] .
Hence, the correct option is option(d).

Note: Substitutions should be done carefully, because any mistake during substitution will lead to wrong answers. Also, the value of the derivative of ${{x}^{x}}$ with respect to $x$ should be remembered as a formula. It is very helpful while solving difficult problems.