Solve (2a)^0/(2a)^5

$$\frac { ( 2 a ) ^ { 0 } } { ( 2 a ) ^ { 5 } }$$

Evaluate

$$\frac{1}{32a^{5}}$$

Solution Steps

Rewrite \left(2a\right)^{5} as \left(2a\right)^{0}\times \left(2a\right)^{5}. Cancel out \left(2a\right)^{0} in both numerator and denominator.

$$\frac{1}{\left(2a\right)^{5}}$$

Expand \left(2a\right)^{5}.

$$\frac{1}{2^{5}a^{5}}$$

Calculate 2 to the power of 5 and get 32.

$$\frac{1}{32a^{5}}$$

Differentiate w.r.t. a

$$-\frac{5}{32a^{6}}$$

Steps Using Chain Rule

Rewrite \left(2a\right)^{5} as \left(2a\right)^{0}\times \left(2a\right)^{5}. Cancel out \left(2a\right)^{0} in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{1}{\left(2a\right)^{5}})$$

Expand \left(2a\right)^{5}.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{1}{2^{5}a^{5}})$$

Calculate 2 to the power of 5 and get 32.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{1}{32a^{5}})$$

If F is the composition of two differentiable functions f\left(u\right) and u=g\left(x\right), that is, if F\left(x\right)=f\left(g\left(x\right)\right), then the derivative of F is the derivative of f with respect to u times the derivative of g with respect to x, that is, \frac{\mathrm{d}}{\mathrm{d}x}(F)\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}x}(f)\left(g\left(x\right)\right)\frac{\mathrm{d}}{\mathrm{d}x}(g)\left(x\right).

$$-\left(32a^{5}\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}a}(32a^{5})$$

The derivative of a polynomial is the sum of the derivatives of its terms. The derivative of a constant term is 0. The derivative of ax^{n} is nax^{n-1}.

$$-\left(32a^{5}\right)^{-2}\times 5\times 32a^{5-1}$$

Simplify.

$$-160a^{4}\times \left(32a^{5}\right)^{-2}$$