How do you find the derivative #e^(x+y) + e^2x +e^2y#?

Question

Emily Ammerman · Accepted Answer

I assume that you want to find #d/dx(e^(x+y)+e^(2x)+e^(2y))#. If I have misunderstood your question, please accept my apology. #d/dx(e^(x+y)+e^(2x)+e^(2y)) = d/dx(e^(x+y))+d/dx(e^(2x))+d/dx(e^(2y))#. For each term, we will need the chain rule applied to the exponential function: #d/dx(e^u) = e^u (du)/dx# #d/dx(e^(x+y)+e^(2x)+e^(2y)) = e^(x+y)d/dx(e^(x+y))+e^(2x)d/dx(2x)+e^(2y)d/dx(2y)#. # = e^(x+y)(1+dy/dx) +2e^(2x)+e^(2y)*2dy/dx# #= e^(x+y)+e^(x+y)dy/dx +2e^(2x)+2e^(2y)dy/dx# So,
#d/dx(e^(x+y)+e^(2x)+e^(2y))=e^(x+y)+e^(x+y)dy/dx +2e^(2x)+2e^(2y)dy/dx# If we had an equation, we could solve for #dy/dx#, but as it is, all we can do is rewrite using algebra: #d/dx(e^(x+y)+e^(2x)+e^(2y))=[e^(x+y)+2e^(2x)]+[e^(x+y) +2e^(2y)]dy/dx#

William Abdalla · Answer

undefined To find the derivative of $ e^{x+y} + e^{2x} + e^{2y} $ with respect to $ x $, you can apply the chain rule and the rule for differentiating exponential functions. The derivative with respect to $ x $ is:

$ \frac{d}{dx} \left( e^{x+y} + e^{2x} + e^{2y} \right) = e^{x+y} \cdot (1 + 1) + e^{2x} \cdot 2 + 0 $

$ = e^{x+y} \cdot 2 + 2e^{2x} $

Similarly, to find the derivative with respect to $ y $, you would get:

$ \frac{d}{dy} \left( e^{x+y} + e^{2x} + e^{2y} \right) = e^{x+y} \cdot (1 + 1) + 0 + e^{2y} \cdot 2 $

$ = e^{x+y} \cdot 2 + 2e^{2y} $