Question

f'(x)=
the equation is below

Answer 1

\[f(x)=6\sqrt{x}(x^3-8\sqrt{x}+7)\]

Answer 2

2) f(t)=(In8)^t

Answer 3

First one: write everthing as powers of x, then expand the brackets and differentiate term-by-term.
So:\[f(x)=6x^{\frac{1}{2}}(x^3-8x^{\frac{1}{2}}+7)=6x^{3\frac{ 1 }{ 2 }} - .........\]

Answer 4

Second one is of the form:\[f(t)=a^t\]so\[f'(t)=\ln a \cdot a^t\]In this function a = ln8.

Answer 5

For the first answer, is it 20x(7/3)-48x+21x^(-1/2)

Answer 6

the second one is In64^t?

Answer 7

First one: \[f(x)=6x^{3.5}-48x+42x^{0.5}\]So \[f'(x)=21x^{2.5}-48+21x^{-0.5}\]Now write with radicals again:\[f'(x)=21x^2\sqrt{x}-48+\frac{ 21 }{ \sqrt{x} }\]

Answer 8

Second one:\[f(t)=(\ln8)^t\]So\[f'(t)=\ln(\ln 8)\cdot (\ln 8)^t\]because a = ln 8 (see my first remark about the derivative of an exponential function).