Solve f(2)^-5*(2)^-7*(2)^8=

$$f ( 2 ) ^ { - 5 } \cdot ( 2 ) ^ { - 7 } \cdot ( 2 ) ^ { 8 } =$$

Evaluate

$$\frac{f}{16}$$

Solution Steps

To multiply powers of the same base, add their exponents. Add -5 and -7 to get -12.

$$f\times 2^{-12}\times 2^{8}$$

To multiply powers of the same base, add their exponents. Add -12 and 8 to get -4.

$$f\times 2^{-4}$$

Calculate 2 to the power of -4 and get \frac{1}{16}.

$$f\times \frac{1}{16}$$

Differentiate w.r.t. f

$$\frac{1}{16} = 0.0625$$

Steps Using Definition of a Derivative

To multiply powers of the same base, add their exponents. Add -5 and -7 to get -12.

$$\frac{\mathrm{d}}{\mathrm{d}f}(f\times 2^{-12}\times 2^{8})$$

To multiply powers of the same base, add their exponents. Add -12 and 8 to get -4.

$$\frac{\mathrm{d}}{\mathrm{d}f}(f\times 2^{-4})$$

Calculate 2 to the power of -4 and get \frac{1}{16}.

$$\frac{\mathrm{d}}{\mathrm{d}f}(f\times \frac{1}{16})$$

The derivative of ax^{n} is nax^{n-1}.

$$\frac{1}{16}f^{1-1}$$

Subtract 1 from 1.

$$\frac{1}{16}f^{0}$$

For any term t except 0, t^{0}=1.

$$\frac{1}{16}\times 1$$

For any term t, t\times 1=t and 1t=t.

$$\frac{1}{16}$$