Question

3) The equation for the instantaneous voltage across a discharging capacitor is given by the equation

𝑦 = 12𝑒 ^-x/2

a) Differentiate the function to find an equation for the gradient and calculate two gradients at x = 2 and x = 4.

4)

𝑦 = 10 𝑙𝑜𝑔3x

a)Differentiate the function to find an equation for the gradient and calculate two gradients at x = 2 and x = 8.

Answer 1

3)

$\frac{dy}{dx}=12\frac{d}{dx}e^{-\frac{x}{2}}\\=12\frac{d}{dx}e^{-\frac{x}{2}}\times \frac{d}{dx}(\frac{-x}{2})$ (chain rule to differentiate on)

$=12\times e^{\frac{-x}{2}}\times \frac{-1}{2}\\\frac{dy}{dx}=-6^{\frac{-x}{2}}$ (equation of the radiant)

at x=2,$\frac{dy}{dx}=-6e^{\frac{-2}{2}}=\frac{-6}{e}\\$

at x=4,$\frac{dy}{dx}=-6e^{\frac{-4}{2}}=\frac{-6}{e^2}\\$

gradient at x=2 is $\frac{-6}{e}\\$

gradient at x=4 is $\frac{-6}{e^2}\\$

4)

$\frac{dy}{dx}=10\frac{d}{dx}log(3x)\\=10\times\frac{1}{ln(10)}\times \frac{1}{3x}\times 3\times 1\\\frac{dy}{dx}=\frac{10}{xln(10)}$ (chain rule to differentiate on)

$\frac{dy}{dx}=\frac{10}{xln(10)}$ (equation of the radiant)

at x=2,$\frac{dy}{dx}=\frac{10}{2ln(10)}=\frac{5}{ln(10)}$

at x=8,$\frac{dy}{dx}=\frac{10}{8ln(10)}=\frac{5}{4ln(10)}$

gradient at x=2 is $\frac{5}{ln(10)}\\$

gradient at x=8 is $\frac{5}{4ln(10)}\\$