How do you differentiate #f(x)=(lnx+x)(tanx+e^x)# using the product rule?

Question

Isaiah Allen · Accepted Answer

#d/dx((lnx+x)(tanx+e^x))=(lnx+x)(e^x+sec^2x)+(tanx+e^x)((tanx+e^x)#, which can be simplified as necessary

Given: #f(x)=(lnx+x)(tanx+e^x)# Let #y=f(x)# if #u=lnx+x#, then #(du)/dx=tanx+e^x# if #v=tanx+e^x#, then #(dv)/dx=sec^2x+e^x# #(dv)/dx=e^x+sec^2x# #f(x)=uv# #y=uv# #(dy)/dx=d/dx(y)# #d/dx(y)=d/dx(uv)# Thus, #(dy)/dx=d/dx(uv)# According to the product rule #d/dx(uv)=u(dv)/dx+v(du)/dx# Substituting #d/dx((lnx+x)(tanx+e^x))=(lnx+x)(e^x+sec^2x)+(tanx+e^x)((tanx+e^x)#

William Abdalla · Answer

undefined To differentiate the function $ f(x) = (ln(x) + x)(tan(x) + e^x) $ using the product rule, you apply the rule which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

First, we identify the two functions:
$ u(x) = ln(x) + x $ and $ v(x) = tan(x) + e^x $

Now, we find the derivatives of each function:
$ u'(x) = \frac{1}{x} + 1 $ and $ v'(x) = sec^2(x) + e^x $

Then, we apply the product rule:
$ f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) $

Substitute the values:
$ f'(x) = (\frac{1}{x} + 1)(tan(x) + e^x) + (ln(x) + x)(sec^2(x) + e^x) $

That's the derivative of the function using the product rule.