Solve (1/2y)^5

$$( \frac { 1 } { 2 y } ) ^ { 5 }$$

Evaluate

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

Solution Steps

To raise \frac{1}{2y} to a power, raise both numerator and denominator to the power and then divide.

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

Calculate 1 to the power of 5 and get 1.

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

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

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

Calculate 2 to the power of 5 and get 32.

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

Differentiate w.r.t. y

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

Steps Using Chain Rule

To raise \frac{1}{2y} to a power, raise both numerator and denominator to the power and then divide.

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

Calculate 1 to the power of 5 and get 1.

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

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

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

Calculate 2 to the power of 5 and get 32.

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

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).

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

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}.

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

Simplify.

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