Answer to Question #260843 in Calculus for Wah

Question #260843

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 π‘™π‘œπ‘”3x

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


1
Expert's answer
2021-11-04T19:24:09-0400

3)

dydx=12ddxeβˆ’x2=12ddxeβˆ’x2Γ—ddx(βˆ’x2)\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)


=12Γ—eβˆ’x2Γ—βˆ’12dydx=βˆ’6βˆ’x2=12\times e^{\frac{-x}{2}}\times \frac{-1}{2}\\\frac{dy}{dx}=-6^{\frac{-x}{2}} (equation of the radiant)


at x=2,dydx=βˆ’6eβˆ’22=βˆ’6e\frac{dy}{dx}=-6e^{\frac{-2}{2}}=\frac{-6}{e}\\

at x=4,dydx=βˆ’6eβˆ’42=βˆ’6e2\frac{dy}{dx}=-6e^{\frac{-4}{2}}=\frac{-6}{e^2}\\

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

gradient at x=4 is βˆ’6e2\frac{-6}{e^2}\\

4)

dydx=10ddxlog(3x)=10Γ—1ln(10)Γ—13xΓ—3Γ—1dydx=10xln(10)\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)


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


at x=2,dydx=102ln(10)=5ln(10)\frac{dy}{dx}=\frac{10}{2ln(10)}=\frac{5}{ln(10)}

at x=8,dydx=108ln(10)=54ln(10)\frac{dy}{dx}=\frac{10}{8ln(10)}=\frac{5}{4ln(10)}

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

gradient at x=8 is 54ln(10)\frac{5}{4ln(10)}\\



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