How do you use the Product Rule to find the derivative of #6e^{9x} sin(6x)#?

Question

Evelyn Agard · Accepted Answer

I found: #y'=18e^(9x)[3sin(6x)+2cos(6x)]# You first derive #e^(9x)# and leave the #sin# as it is then you add the first part as it is and derive #sin# (to derive the arguments of the two functions you may use the Chai Rule): So, you get: #y'=6e^(9x)*9*sin(6x)color(red)(+)6e^(9x)cos(6x)*6=# #=54e^(9x)sin(6x)+36e^(9x)cos(6x)=# #=9e^(9x)[6sin(6x)+4cos(6x)]# #=18e^(9x)[3sin(6x)+2cos(6x)]#

Ezra Adams · Answer

undefined To find the derivative of $6e^{9x} \sin(6x)$ using the Product Rule, we differentiate each term separately and then apply the rule. The Product Rule states that if $u(x)$ and $v(x)$ are differentiable functions of $x$, then the derivative of their product $u(x)v(x)$ is given by $u'(x)v(x) + u(x)v'(x)$.

Let $u(x) = 6e^{9x}$ and $v(x) = \sin(6x)$.

Then, $u'(x) = 6 \cdot 9e^{9x} = 54e^{9x}$ (using the chain rule for the exponential function) and $v'(x) = 6\cos(6x)$ (using the derivative of sine function).

Now, applying the Product Rule, we have:

$$\frac{d}{dx} (u(x)v(x)) = u'(x)v(x) + u(x)v'(x)$$

$$\frac{d}{dx} (6e^{9x} \sin(6x)) = (54e^{9x})\sin(6x) + 6e^{9x}\cos(6x)$$

So, the derivative of $6e^{9x} \sin(6x)$ with respect to $x$ is $54e^{9x}\sin(6x) + 6e^{9x}\cos(6x)$.