Differentiate formula for amount of charge on capacitor

[Hassss]

Hi

Ive got some calculations to do on amount of charge on a capacitor with respect to t.

Formula for amount of charge on a capacitor is Q=5.7ln4.5t

I need to determine the rate of discharge with different values of t, t=5ms t=7.5ms

I know i need to differentiate the formula against time as it will give me the rate of charge.

Ive completely forgotten how to apply calculus and would sincerely appreciate any help and advice!

thank you

 

[Matterwave]

Do you know what $\frac{d}{dx}\ln (x)$ is? How much of calculus have you forgotten? Do you know the chain rule?

 

[Hassss]

Do you know what $\frac{d}{dx}\ln (x)$ is? How much of calculus have you forgotten? Do you know the chain rule?

Hi

the last time i did calculus was approx 1 1/2 years ago I've literally have no clue what the chain rule or the relationship you've stated means! :(

 

[Matterwave]

Hmm, that might make this kind of difficult!

Well, the chain rule says, given functions f(x) and g(y) if I have a composite function g(f(x)) then the derivative with respect to x is:

$$\frac{d}{dx}g(f(x))=\frac{dg}{df}\frac{df}{dx}$$

As an example, if I have, say $\sin(5x)$ and I want to take the derivative with respect to x, I know that $\frac{d}{dx}\sin(x)=\cos(x)$ and so if I take g(y)=sin(y) and f(x)=5x, then by the chain rule I can find:

$$\frac{d}{dx}\sin(5x)=\frac{d}{dx}\sin(f)=\cos(f)\frac{df}{dx}=5\cos(5x)$$

Also, note that the ln is the natural logarithm defined such that:

$$\int_1^x \frac{1}{y}dy=\ln(x)$$

By the fundamental theorem of calculus, then we know:

$$\frac{d}{dx}\ln(x)=\frac{1}{x}$$

Can you use this information to solve your problem?