Question

2.3 Find the derivative of
f(x) =x2x+3e 3x+e2xcosx when x is:
A) 3
B) 2
C) 0

Answer 1

The function given to us is:

$f(x)=x^{2x}+3e^{3x}+e^{2x}cosx$

This can be broken down to three constituent functions as:

$y_1=x^{2x}$, $y_2=3e^{3x}$ and $y_3=e^{2x}cosx$. Now, the derivative of the function f(x) will be the sum of the derivatives of these individual functions $y_1, y_2, y_3$.

The first one is the toughest and it can be found by taking natural log on both sides and then taking the derivative as:

$ln(y_1)=2xln(x)$

Taking derivative we get:

$\frac{1}{y_1}y_1'= 2lnx+\frac{2x}{x} =2lnx+2$

$\therefore y_1'=y_1(2lnx+2)=2x^{2x}(lnx+1)$...........Equation 1

Likewise we can easily find the derivatives of $y_2, y_3$ as:

$y_2=9e^{3x}$ .......................Equation 2

$y_3=e^{2x}(2cos(x)-sin(x))$ .......Equation 3

Thus, the derivative of f(x) will be

$f'(x)=2x^{2x}(lnx+1)+9e^{3x}+e^{2x}(2cos(x)-sin(x))$

Now, plugging in we get:

a) At x=3, f'(x)=75131.8

b) At x=1, f'(x)=184.54

c) Using Limits, as x->0, we get, at x=0, $f'(x)=-\infty$