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Question

Question: If we have an expression as \({{x}^{m{{x}^{m{{x}^{m{{x}^{mx\ldots }}}}}}}}={{y}^{n{{y}^{n{{y}^{n{{y}...

If we have an expression as xmxmxmxmx=ynynynyny{{x}^{m{{x}^{m{{x}^{m{{x}^{mx\ldots }}}}}}}}={{y}^{n{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}}, then dydx\dfrac{dy}{dx} is equal to:
1. yx\dfrac{y}{x}
2. xy\dfrac{x}{y}
3. mymx\dfrac{my}{mx}
4. nymx\dfrac{ny}{mx}

Explanation

Solution

For solving this question you should know about the differentiation and differentiation of a function to infinite power. In this problem first we will put the function as it is given and then make it equal to one by dividing sufficient terms on both sides and then shorten this as much as possible. And then at last we will differentiate that function to get the answer.

Complete step-by-step solution:
According to the question the function is given to us as xmxmxmxmx=ynynynyny{{x}^{m{{x}^{m{{x}^{m{{x}^{mx\ldots }}}}}}}}={{y}^{n{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}}and asked to find dydx\dfrac{dy}{dx} for this. So, as we can see here that the function is given as,
xm(xmxmxmx)=ynynynyny{{x}^{m\left( {{x}^{m{{x}^{m{{x}^{mx\ldots }}}}}} \right)}}={{y}^{n{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}}
As we can see that the power is going till infinite, so we can reduce xmxmxmxmx=ynynynyny{{x}^{m{{x}^{m{{x}^{m{{x}^{mx\ldots }}}}}}}}={{y}^{n{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}} to our expression and we can write it as,
xm(ynynyny)=ynynynyny{{x}^{m\left( {{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}} \right)}}={{y}^{n{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}}
Now for making it’s value 1, we divide both the sides by ynynynyny{{y}^{n{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}}.
So, we get
xmynynynyynynynyny=ynynynynyynynynyny (xm)ynynyny(yn)ynynyny=1 \begin{aligned} & \dfrac{{{x}^{m{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}}}{{{y}^{n{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}}}=\dfrac{{{y}^{n{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}}}{{{y}^{n{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}}} \\\ & \Rightarrow \dfrac{{{\left( {{x}^{m}} \right)}^{{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}}}{{{\left( {{y}^{n}} \right)}^{{{y}^{n{{y}^{n{{y}^{ny\ldots }}}}}}}}}=1 \\\ \end{aligned}
Now it can be further written as,
(xmyn)ynyny=1{{\left( \dfrac{{{x}^{m}}}{{{y}^{n}}} \right)}^{{{y}^{n{{y}^{ny\ldots }}}}}}=1
And here we can easily remove this power, because we know that 1 to the power anything is always 1. So, we get
xmyn=1\dfrac{{{x}^{m}}}{{{y}^{n}}}=1
We now multiply both the sides by yn{{y}^{n}}, so we will get,
xm=yn{{x}^{m}}={{y}^{n}}
And now differentiate the above equation with respect to x, so we will get as follows,
mxm1=nyn1.dydx dydx=(mn)xm1yn1 \begin{aligned} & m{{x}^{m-1}}=n{{y}^{n-1}}.\dfrac{dy}{dx} \\\ & \Rightarrow \dfrac{dy}{dx}=\left( \dfrac{m}{n} \right)\dfrac{{{x}^{m-1}}}{{{y}^{n-1}}} \\\ \end{aligned}
So, this is the final value of dydx\dfrac{dy}{dx}.

Note: For solving these types of questions you have to know if the power of any variable is going till infinite, then always reduce that by making any mathematical operation of division or multiplication.