Question

find the derivative function of e^2x log(3x+4) w.r.t x​

Answer 1

Step-by-step explanation:

e^2 x (2) log(3x+4) + e^2 x (1/3 x+4) (3)

please mark as brillianist answer

Answer 2

The derivative of the given function of x is $e^{2x}(\frac{3}{3x+4} + 2log(3x+4))$.

.Here, the function "y" that we want to find the derivative of is a product of two functions of functions:

The first function is a composite of exponential i.e. $e^{2x}$ and product function i.e. $2x$.

The second function is a composite of the logarithm function i.e. $log(x)$

outer function i.e. $3x+4$.

We first individually find the derivative of each function independently using the chain rule.

Function 1 : Exponential and Product Function

The first part is : $e^{2x}$

The differentiation of this function using the chain rule is :

             $dy/dx = d(e^{2x} )/dx\\\\dy/dx = d(e^{2x} )/d(2x) * d(2x)/dx\\\\dy/dx = (e^{2x}) * (2)\\\\dy/dx = 2e^{2x}$

Function 2 : Logarithm and Product Function

The second part is : $log(3x+4)$

The differentiation of this function using the chain rule is :

              $dy/dx = d(log(3x+4))/dx\\\\dy/dx = d(log(3x+4))/d(3x+4) *d(3x+4)/dx\\\\dy/dx = (1/(3x+4) ) *(3)\\\\dy/dx = 3/(3x+4)$

             

Now, for the differentiation of the product of these two parts, we use the product rule :

       $d(a*b)/dx = (a*db/dx) + (b*da/dx)$

Thus, applying the product rule :

      $d(e^{2x}log(3x+4))/dx = ((e^{2x})*d(log(3x+4))/dx) + ((log(3x+4))*d(e^{2x})/dx)\\\\d(e^{2x}log(3x+4))/dx = ((e^{2x})*(3/(3x+4)) + ((log(3x+4)*(2e^{2x}))\\\\d(e^{2x}log(3x+4))/dx = (3e^{2x}/(3x+4)) + (2e^{2x}log(3x+4))\\\\d(e^{2x}log(3x+4))/dx = e^{2x}(\frac{3}{3x+4} + 2log(3x+4))$

       

Hence, the derivative of the given function of 'x' is $e^{2x}(\frac{3}{3x+4} + 2log(3x+4))$

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