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How do you differentiate # y =e^(2x )(x^2+5^x)# using the chain rule?

Answer 1

Samantha Adkison

#y'=e^(2x)(2x^2+2(5^x)+2x+5^xln5)#

Through the product rule:

#y'=(x^2+5^x)color(blue)(d/dx[e^(2x)])+e^(2x)color(magenta)(d/dx[x^2+5^x])#

Find each derivative individually (both will require the chain rule):

#color(blue)(d/dx[e^(2x)])=e^(2x)d/dx[2x]=color(blue)(2e^(2x))#

#color(magenta)(d/dx[x^2+5^x])=2x+color(green)(d/dx[5^x]#

Finding #d/dx[5^x]# will be much simpler if we recognize that #5^x=e^(ln5^x)=e^(xln5)#.

#color(green)(d/dx[e^(xln5)])=overbrace(e^(xln5))^("still"=5^x)d/dx[xln5]=color(green)(5^xln5#

So, plug this back into the derivative we were finding earlier:

#d/dx[x^2+5^x]=color(magenta)(2x+5^xln5#

Plug everything we know back into our original equation for #y'#:

#y'=color(blue)(2e^(2x))(x^2+5^x)+e^(2x)(color(magenta)(2x+5^xln5))#

#y'=e^(2x)(2x^2+2(5^x)+2x+5^xln5)#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.