What is the derivative of #(x^2)(e^(4x))#?

Question

Christopher Allers · Accepted Answer

#2xe^(4x)(2x+1)# suppose, #f(x)=x^2e^(4x)# #d/(dx)(f(x))# #=d/(dx)(x^2e^(4x))# #=x^2d/(dx)(e^(4x))+e^(4x)d/(dx)(x^2)# #=x^2e^(4x)d/(dx)(4x)+2xe^(4x)# #=4x^2e^(4x)+2xe^(4x)# #=2xe^(4x)(2x+1)#

Luke Alvarez · Answer

# 2xe^(4x) ( 2x + 1 ) #

Differentiate # color(blue)(" using product rule ") # # d/dx( x^2 e^(4x)) = x^2 d/dx(e^(4x)) + e^(4x) d/dx(x^2) # # = x^2(e^(4x) d/dx(4x)) + e^(4x)(2x) # # = x^2. e^(4x) .4 + e^(4x) .2x # # = 4x^2e^(4x) + 2xe^(4x) = 2xe^(4x)(2x + 1 ) #

Ezra Adams · Answer

#2xe^(4x)(2x+1)#

Use the product rule, which states that for a function #y=f(x)g(x)#,

#y'=f'(x)g(x)+g'(x)f(x)#

This can also be expressed in English as follows: the derivative of two functions multiplied by one another is equal to the derivative of one times the nondifferentiated form of the other plus the derivative of the other function times the differentiated form in the other term.

This is what we have

#f(x)=x^2# #g(x)=e^(4x)#

Thus, the derivatives of them are

#f'(x)=2x# #g'(x)=4e^(4x)#

Note that finding #g'(x)# require the chain rule. The chain rule states that for a function #y=h(k(x))#,

#y'=h'(k(x))*k'(x)#

In the case of #e^(4x)#,

#h(x)=e^x# #k(x)=4x#

#h'(x)=e^x# #k'(x)=4#

When we enter these values into the chain rule, we can see that

#d/dx(e^(4x))=e^(4x)*4=4e^(4x)#

When we go back to the prior set of derivatives and functions, we can see that

#d/dx(x^2e^(4x))=2xe^(4x)+4x^2e^(4x)#

This can be made simpler as

#=2xe^(4x)(2x+1)#

Evelyn Agard · Answer

undefined To find the derivative of the function $ f(x) = x^2 \cdot e^{4x} $, you can use the product rule of differentiation. The product rule states that if you have two functions, $ u(x) $ and $ v(x) $, then the derivative of their product is $ u'(x) \cdot v(x) + u(x) \cdot v'(x) $.

So, for $ f(x) = x^2 \cdot e^{4x} $, let $ u(x) = x^2 $ and $ v(x) = e^{4x} $. Then, $ u'(x) = 2x $ and $ v'(x) = 4e^{4x} $.

Applying the product rule, we have:

$$ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) = (2x) \cdot e^{4x} + x^2 \cdot (4e^{4x}) $$

Therefore, the derivative of $ f(x) = x^2 \cdot e^{4x} $ is:

$$ f'(x) = 2x \cdot e^{4x} + 4x^2 \cdot e^{4x} $$