Use logarithmic differentiation

Question

Determine q′(x) if q(x)=(3x³+4x)^cosx. Enter an exact answer in terms of a natural logarithm.

Answer 1

Hi Henry,

Start by taking the natural of each side.

ln(q) = ln(3x^3+4x)^cos(x)

Using the fact that ln(A)^B = B*ln(A), rearrange the right side of the equation.

ln(q) = cos(x) * ln(3x^3+4x)

Implicitly derive both sides of the equation with d/dx. The right side of the equation is a product rule with a chain rule for the logarithm.

d/dx(ln(q)) = d/dx(cos(x) * ln(3x^3+4x))

1/q*dq/dx = (-sin(x))*ln(3x^3+4x) + cos(x)*1/(3x^3+4x)*(9x^2+4)

1/q*dq/dx = (-sin(x))*ln(3x^3+4x) + cos(x)*(9x^2+4)/(3x^3+4x)

There really isn't a whole lot to be gained by simplifying the right side of the equation.

Multiply both sides by q to isolate dq/dx

dq/dt = q*(-sin(x))*ln(3x^3+4x) + cos(x)*(9x^2+4)/(3x^3+4x)

q is just your original function. Replace q with your first equation to get your incredibly ugly final answer.

dq/dt = (3x^3+4x)^cos(x)*(-sin(x))*ln(3x^3+4x) + cos(x)*(9x^2+4)/(3x^3+4x)

Hope this helps.