Second Derivative of an equation: The charge ? on the plate of a capacitor is given by ? = ???^(− ?/??)...

[Core7551]

The charge ? on the plate of a capacitor is given by ? = ???^(− ?/??),
where ? is the time, ? is the capacitance and ? is the resistance.
Suggest a solution to determine the second derivative. I'd like to know why e^(-t/CR) remains the same throughout the equation

 

[stapel]

The charge ? on the plate of a capacitor is given by ? = ???^(− ?/??),
where ? is the time, ? is the capacitance and ? is the resistance.
Suggest a solution to determine the second derivative. I'd like to know why e^(-t/CR) remains the same throughout the equation

What do you mean by "throughout the equation"? (I mean, it's not like the equation is supposed to change itself somehow. The derivative, of course, is based on that equation, but is itself a different equation.)

 

[topsquark]

The charge ? on the plate of a capacitor is given by ? = ???^(− ?/??),
where ? is the time, ? is the capacitance and ? is the resistance.
Suggest a solution to determine the second derivative. I'd like to know why e^(-t/CR) remains the same throughout the equation

Think about it.
[imath]f(x) = Ae^{bx}[/imath]

By the chain rule
[imath]f^{\prime}(x) = A \dfrac{d ( e^{bx} )}{d(bx)} \cdot \dfrac{d(bx)}{dx} = A e^{bx} b = Ab e^{bx}[/imath]

Rinse and repeat.

If you want to know why [imath]\dfrac{d (e^x)}{dx} = e^x[/imath], then see here. (Wait a moment for the LaTeX to render.)

-Dan

 

[Core7551]

I solved the question and got this equation for the first derivative:
[math]\frac {dq}{dt}=\frac {-V}{R}e^\frac{-t}{CR}[/math]
My lecturer is saying the answer is :
[math]\frac{dq}{dt}=\frac{-V}{R} e^\frac{-t}{CR}-1[/math].

I'd like to understand his answer better. Please help.

 

[stapel]

I solved the question and got this equation for the first derivative:
[math]\frac {dq}{dt}=\frac {-V}{R}e^\frac{-t}{CR}[/math]
My lecturer is saying the answer is :
[math]\frac{dq}{dt}=\frac{-V}{R} e^\frac{-t}{CR}-1[/math].

I'd like to understand his answer better. Please help.

I don't get his answer. Neither does Wolfram Alpha.

 

[Core7551]

I don't get his answer. Neither does Wolfram Alpha.

He is using the power rule; the he is subtracting 1 from the power .

 

[stapel]

He is using the power rule; the he is subtracting 1 from the power .

If he were subtracting 1 from the power, then he would have put the "-1" in the power, rather than subtracting it from the derivative (as you've shown). Also, if that's what he's doing, then he doesn't know how to differentiate exponentials.

 

[Core7551]

That was my bad. Sorry about that. Can you please give me a simple explanation about differentiating exponential?

 

[stapel]

That was my bad. Sorry about that. Can you please give me a simple explanation about differentiating exponential?

It is not reasonably feasible to attempt here to replace the section or chapter in your textbook, nor the days of instruction in the classroom. Fortunately, there are loads of lessons available online. [list]

 

[Dr.Peterson]

I solved the question and got this equation for the first derivative:
[math]\frac {dq}{dt}=\frac {-V}{R}e^\frac{-t}{CR}[/math]
My lecturer is saying the answer is :
[math]\frac{dq}{dt}=\frac{-V}{R} e^\frac{-t}{CR}-1[/math].

I'd like to understand his answer better. Please help.

He is using the power rule; the he is subtracting 1 from the power .

That was my bad. Sorry about that. Can you please give me a simple explanation about differentiating exponential?

Since you got the correct first derivative, and your lecturer did it wrong, I suppose you want to give that lesson to him?

Here is one:

6. Derivative of the Exponential Function

How to differentiate exponential functions, with examples.

www.intmath.com

The power rule applies when the base is a variable and the exponent is a constant; here the base is a constant and the exponent is a variable!