Question

How do you find the derivative of y = \[{{e}^{ex}}\ln (x)\]?

Answer 1

Hint: To solve the above given question we will be using the multiplication rule as we can see that y consists of two consecutive functions i.e. f(x) = g(x) h(x) and to solve this type of functions we can use the multiplication property of derivatives.

Complete step by step solution:
Let us say y = f(x)
Now with the use of multiplication rule, in which it is, clearly stated that,
If f(x) = g(x) h(x)
Then f’(x) = g’(x) h(x) + g(x) h’(x)…….. (1)
Where
⇒ f’(x) = \[\dfrac{df}{dx}\] ( this defines the derivative of f(x))
⇒g’(x) = \[\dfrac{dg}{dx}\] (this defines the derivative of g(x))
⇒h’(x) = \[\dfrac{dh}{dx}\] (this defines the derivative of h(x))
So in the question that is stated above, the values that are required to solve the question are stated as under:
⇒f(x) = \[{{e}^{ex}}\ln (x)\]
⇒g(x) = \[{{e}^{ex}}\]
⇒h(x) = ln(x)
Now, to calculate the derivative of the functions we will have to know the derivative of some values: So, those values for which derivatives are required are stated as below:
Differentiation of \[{{e}^{ax}}=\text{ }a{{e}^{ax}}\]
Differentiation of ln(x) = \[\dfrac{1}{x}\]
By using all the above stated property and then applying it in those functions for which we need differentiated value and then substituting it in equation (1) we can get:
⇒f’(x) = \[e{{e}^{ex}}\ln (x)\]+ \[\dfrac{e{{e}^{ex}}}{x}\]
So, after substituting the derivatives of function in equation (1) we can clearly say that the derivative of
y = \[{{e}^{ex}}\ln (x)\] is \[e{{e}^{ex}}\ln (x)\]+ \[\dfrac{e{{e}^{ex}}}{x}\]

Note:
The point where mistakes are generally made is to how the multiplication rule should be applied. We can apply the multiplication rule in the format F(x) = g(x)h(x)c(x) which will give this required derivative of the function as F’(x) = g’(x)h(x)c(x) + g(x)h’(x)c(x) + g(x)h(x)c’(x)
In the above mentioned method there were three functions, which needed to be solved so, we first differentiate the 1st function and then leave the 2nd and 3rd function as it is and then we differentiate the 2nd function and leave the 1st and 3rd function as it is and then for the last part we differentiate the 3rd function and leave the 1st and 2nd function as it is. In the multiplication rule for each part we will differentiate only 1 function in ascending order and leave the remaining functions as it is. We will not differentiate those functions which we have already differentiated before.