Use logarithmic differentiation to find the derivative of y.
y = x(x + 8) (x + 9)

y'=3x^2+34x+72

Step-by-step explanation:

I see this has no answer and it might be too late.

It's asking us to find dy/dx, after taking log of both sides and apply any useful properties of log.

Take log of both sides:

log(y)=log[x(x+8)(x+9)]

Apply product rule of logarithms on right:

log(y)=log(x)+log(x+8)+log(x+9)

Now differentiate both sides:

y'/y=1/x+1/(x+8)+1/(x+9)

Multiply both sides by y:

y'=y(1/x+1/(x+8)+1/(x+9))

Replace y with x(x+8)(x+9) -> this is from the given equation:

y'=x(x+8)(x+9)(1/x+1/(x+8)+1/(x+9))

Distribute:

y'=(x+8)(x+9)+x(x+9)+x(x+8)

Multiply/distribute some more:

y'=x^2+17x+72+x^2+9x+x^2+8x

Combine like terms:

y'=3x^2+34x+72

He burnt 1000 calories per hour when playing basketball.

Step-by-step explanation:

Let B be calories burned playing basketball, and C calories burned canoing.

1800 = B + 2C

3200 = 2B + 3C

From 1st equatipn, we get that B = 1800 - 2C

Replacing into the 2nd equation, we have:

3200 = 2(1800-2C) + 3C

3200 = 3600 - 4C + 3C

3200 = 3600 - 1C

C = 3600 - 3200

C = 400

Knowing C, we find B.

B = 1800 - 2C = 1800 - 2*400 = 1800 - 800 = 1000 calories.