Need help with a few Calculus problems?

1)

since variable x is in both base and exponent you have to use logarithmic differentiation

let y = x^(3x)

taking logs

ln(y) = 3x ln(x)

differentiating with respect to x

(1/y)dy/dx = 3x(1/x) + ln(x)*3

dy/dx = 3y[1 + ln(x)]

substituting y = x^(3x)

dy/dx = 3x^(3x)[1 + ln(x)]

2)

let y = 2ln[2x + 4ln(x)]

dy/dx = 2/ln[2x + 4ln(x)]*[2 + 4/x]

dy/dx = 2(1/ln(2x+4ln(x))(2x + 4)/x

dy/dx = [4(x+2)/x)](1 /ln(2x+4ln(x))

3)

draw the triangle ABC with B is in the ground. Draw ED perpendicular from AC to BC to represent womans height.

AB = pole height = 14

ED = 6

let BC = y and BD = x, so DC = y - x

now ABC and EDC are similar triangles

so y/(y-x) = 14/6

crossmultiply

6y = 14(y - x)

6y = 14y - 14x

8y = 14x

divide by 2

4y = 7x

differentiating with t

4dy/dt = 7dx/dt

substitute dx/dt = 5 ft/s

4dy/dt = 7*5 = 35

dy/dt = 35/4 = 8.75 ft/s

so shadow is moving at the rate of 8.75 ft/s

f(x)=e^[2x-6] ln[f(x)] = ln[e^(2x-6)] ln[f(x)] = 2x - 6 d/dx ln[f(x)] = d/dx (2x - 6) f'(x)/f(x) = 2 f'(x) = 2e^[2x-6] or with the chain rule in case you like, f'(x) = d/du e^u du/dx (2x - 6) = 2e^[2x-6] The slope of the tangent line at (3,a million) is f'(3) = 2e^0 = 2 answer (a): 2 i think of a few thing is lacking in part b because of the fact (0,a million) does not fulfill f(x) = e^(2x - 6) so which you would be able to not have a tangent to the curve at that factor.

1. LOG IS THE INVERSE OF AN EXPONENT