How do you differentiate #y=ae^x+b/v+c/v^2#?

Answer 1

I will assume that we need to find #dy/dx# and that #a#, #b#, and #c# are constants and that #v# is some unspecified function of #x#.

#y=ae^x+bv^-1 + cv^-2#

#dy/dx = ae^x-bv^(-1-1) (dv)/dx -2c v^(-2-1) (dv)/dx#

# = ae^x-bv^-2 (dv)/dx -2c v^-3 (dv)/dx#

# = ae^x - b/v^2 (dv)/dx - (2c)/v^3 (dv)/dx#

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Answer 2

Luke Alvarez

To differentiate ( y = ae^x + \frac{b}{\sqrt{v}} + \frac{c}{v^2} ), you can use the following steps:

Differentiate each term separately using the rules of differentiation.
For ( ae^x ), the derivative is ( ae^x ).
For ( \frac{b}{\sqrt{v}} ), apply the power rule for differentiation. The derivative is ( -\frac{b}{2v^{3/2}} ).
For ( \frac{c}{v^2} ), apply the power rule again. The derivative is ( -\frac{2c}{v^3} ).
Sum up the derivatives of each term to get the final result.

So, the derivative of ( y ) with respect to ( x ) is:

[ \frac{dy}{dx} = ae^x - \frac{b}{2v^{3/2}} - \frac{2c}{v^3} ]

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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.