Dirivitive using the power "e"

**skyslimit** · Sep 27th 2008, 01:56 PM

Can anyone help me with a few problems please?

I'm not sure how to find the derivative of a function like this.. $5x^e+15e^x$

also, what is the difference between (d/dt) and (d/dx). I found on (d/dt) that I can solve normally, however (d/dx) asks for a different answer?

**Jhevon** · Sep 27th 2008, 02:49 PM

Originally Posted by skyslimit

Can anyone help me with a few problems please?

I'm not sure how to find the derivative of a function like this.. $5x^e+15e^x$

$e$ is a constant. so for $x^e$ just use the power rule to differentiate it. as for $e^x$ its derivative is itself.

also, what is the difference between (d/dt) and (d/dx). I found on (d/dt) that I can solve normally, however (d/dx) asks for a different answer?

i suppose you are talking in the context of implicit differentiation?

anyway, the notation is interpreted thus: $\frac d{dx}$ mean you are taking the derivative with respect to $x$ . an analogous interpretation holds for $\frac d{dt}$

example:

differentiate implicitly with respect to $x$

$x^2 + y^2 = 2x$

$\Rightarrow 2x~\frac {dx}{dx} + 2y~\frac {dy}{dx} = 2~\frac {dx}{dx}$

so we have

$2x + 2y~\frac {dy}{dx} = 2$

(we put dx/dx to say we took the "derivative of an x-term with respect to x" and dy/dx to say we took the "derivative of a y-term with respect to x")

if we differentiated the same equation with respect to $t$ we would get

$2x~\frac {dx}{dt} + 2y~\frac {dy}{dt} = 2~\frac {dx}{dt}$

(we put dx/dt to say we took the "derivative of an x-term with respect to t" and dy/dt to say we took the "derivative of a y-term with respect to t")

see the difference?

**skyslimit** · Sep 27th 2008, 03:51 PM

Thank you.

If I'm not mistaken, are you saying the only difference of the two equations are the variable. So i should be able to find the derivitive of a similar eqaution and find the same answer?

**Jhevon** · Sep 27th 2008, 04:14 PM

Originally Posted by skyslimit

Thank you.

If I'm not mistaken, are you saying the only difference of the two equations are the variable. So i should be able to find the derivitive of a similar eqaution and find the same answer?

the difference is the variable, but that difference can have a lot of repercussions. in general, the answer would not be the same. and it depends on what you are solving for. there are more unknowns in the latter than the former

**skyslimit** · Sep 27th 2008, 04:40 PM

would you mind showing how i can solve for d/dx using an equation like $e^(3x^2+5x)$

**Jhevon** · Sep 27th 2008, 04:43 PM

Originally Posted by skyslimit

would you mind showing how i can solve for d/dx using an equation like $e^(3x^2+5x)$

recall, $\frac d{dx} e^u = u' e^u$ , where $u$ is a function of $x$

thus, $\frac d{dx} e^{3x^2 + 5x} = \bigg[ \frac d{dx}(3x^2 + 5x) \bigg] \cdot e^{3x^2 + 5x}$

can you continue?

and next time, post a new question in a new thread

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