Question

Find dy, dx if f(x) = (x + 8)3x. 3xln(x + 8) the product of the quantity 3 times the natural log of the quantity x plus 8 plus 3 times x divided by the quantity x plus 8, and the quantity x plus 8 raised to the 3x power 3 times the natural log of the quantity x plus 8 plus 3 times x divided by the quantity x plus 8 3x(x + 8)(3x − 1)

Answer 1

Answer:

This is quite a complex problem. I wrote out a really nice solution but I can't work out how to put it on the website as the app is very poorly made. Still, I'll just have to type it all in...Okay so you need to use a technique called logarithmic differentiation. It seems quite unnatural to start with but the result is very impressive. Let y = (x+8)^(3x)Take the natural log of both sides:ln(y) = ln((x+8)^(3x))By laws of logarithms, this can be rearranged:ln(y) = 3xln(x+8)Next, differentiate both sides. By implicit differentiation:d/dx(ln(y)) = 1/y dy/dxThe right hand side is harder to differentiate. Using the substitution u = 3x and v = ln(x+8):d/dx(3xln(x+8)) = d/dx(uv)du/dx = 3Finding dv/dx is harder, and involves the chain rule. Let a = x+ 8:v = ln(a)da/dx = 1dv/da = 1/aBy chain rule:dv/dx = dv/da * da/dx = 1/a = 1/(x+8)Finally, use the product rule:d/dx(uv) = u * dv/dx + v * du/dx = 3x/(x+8) + 3ln(x+8)This overall produces the equation:1/y * dy/dx = 3x/(x+8) + 3ln(x+8)We want to solve for dy/dx, achievable by multiplying both sides by y:dy/dx = y(3x/(x+8) + 3ln(x+8)) Since we know y = (x+8)^(3x):dy/dx = ((x+8)^(3x))(3x/(x+8) + 3ln(x+8))Neatening this up a bit, we factorise out 3/(x+8):dy/dx = (3(x+8)^(3x-1))(x + (x+8)ln(x+8))Well wasn't that a marathon? It's a nightmare typing that in, I hope you can follow all the steps.I hope this helped you :)

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