Question

How do you find \[\dfrac{dy}{dx}\] by implicit differentiation given \[{{e}^{x}}=\ln y\]?

Answer 1

Hint: Differentiate both sides of the given relation with respect to the variable x. Use the formula, \[\dfrac{d\left( {{e}^{x}} \right)}{dx}={{e}^{x}}\] to evaluate the L.H.S. Now, for the R.H.S. use the chain rule of differentiation. First differentiation \[\ln y\] with respect to y and then differentiate y with respect to x. Leave the term \[\dfrac{dy}{dx}\] at one side and take the other terms to the other side and simplify to get the answer. Use the formula: - \[{{\log }_{a}}b=m\], then \[b={{a}^{m}}\], to substitute the value of y in terms of x.

Complete step by step answer:
Here, we have been provided with the relation: - \[{{e}^{x}}=\ln y\] and we are asked to find the value of \[\dfrac{dy}{dx}\].
Now, differentiating both sides of the relation with respect to the variable x, we get,
\[\Rightarrow \dfrac{d\left( {{e}^{x}} \right)}{dx}=\dfrac{d\left( \ln y \right)}{dx}\]
We know that the derivative of exponential function of \[{{e}^{x}}\] is \[{{e}^{x}}\] only, so we have,
\[\Rightarrow {{e}^{x}}=\dfrac{d\left( \ln y \right)}{dx}\]
Using chain rule of differentiation in the R.H.S, we have,
\[\Rightarrow {{e}^{x}}=\dfrac{d\left( \ln y \right)}{dy}\times \dfrac{dy}{dx}\]
What we are doing is, first we are differentiating \[\ln y\] with respect to y and then we are differentiating y with respect to x, and their product is considered. So, we have using the relation \[\dfrac{d\ln x}{dx}=\dfrac{1}{x}\], we get,
\[\Rightarrow {{e}^{x}}=\dfrac{1}{y}\times \dfrac{dy}{dx}\]
\[\Rightarrow \dfrac{dy}{dx}=y{{e}^{x}}\] - (1)
Now, let us come back to the initial relation, we have,
\[\Rightarrow {{e}^{x}}=\ln y\]
 ‘\[\ln \]’ is natural log, i.e., log to the base ‘e’, so we have,
\[\Rightarrow {{e}^{x}}={{\log }_{e}}y\]
Applying the formula: - if \[{{\log }_{a}}b=m\], then \[b={{a}^{m}}\], i.e., conversion of logarithmic function into exponential function, we get,
\[\begin{align}
  & \Rightarrow {{e}^{x}}=y \\
 & \Rightarrow y={{e}^{{{e}^{x}}}} \\
\end{align}\]
Substituting the value of y in equation (1), we get,
\[\Rightarrow \dfrac{dy}{dx}={{e}^{{{e}^{x}}}}.{{e}^{x}}\]
Using the formula: - \[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}\], we get,
\[\Rightarrow \dfrac{dy}{dx}={{e}^{\left( {{e}^{x}}+x \right)}}\]
Hence, the above relation is our answer.



Note:
 One may note that we can also solve the question in a different manner. We can apply the conversion formula of logarithmic function into exponential function at the initial stage of the solution. Then we will assume \[{{e}^{{{e}^{x}}}}\] as \[{{e}^{f\left( x \right)}}\] and differentiate both the sides with respect to x, using the formula, \[\Rightarrow \dfrac{d\left[ {{e}^{f\left( x \right)}} \right]}{dx}={{e}^{f\left( x \right)}}.f'\left( x \right)\], we will get the answer. You must remember the different rules of differentiation like: - the product rule, chain rule, \[\dfrac{u}{v}\] rule etc. Because these are common rules and are used everywhere in calculus.