Solve x/204/3a^2x=

$$\frac { x } { 20 } \frac { 4 } { 3 a ^ { 2 } x } =$$

Evaluate

$$\frac{1}{15a^{2}}$$

Solution Steps

Multiply \frac{x}{20} times \frac{4}{3a^{2}x} by multiplying numerator times numerator and denominator times denominator.

$$\frac{x\times 4}{20\times 3a^{2}x}$$

Cancel out 4x in both numerator and denominator.

$$\frac{1}{3\times 5a^{2}}$$

Multiply 3 and 5 to get 15.

$$\frac{1}{15a^{2}}$$

Differentiate w.r.t. a

$$-\frac{2}{15a^{3}}$$

Steps Using Chain Rule

Multiply \frac{x}{20} times \frac{4}{3a^{2}x} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{x\times 4}{20\times 3a^{2}x})$$

Cancel out 4x in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{1}{3\times 5a^{2}})$$

Multiply 3 and 5 to get 15.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{1}{15a^{2}})$$

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).

$$-\left(15a^{2}\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}a}(15a^{2})$$

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}.

$$-\left(15a^{2}\right)^{-2}\times 2\times 15a^{2-1}$$

Simplify.

$$-30a^{1}\times \left(15a^{2}\right)^{-2}$$

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

$$-30a\times \left(15a^{2}\right)^{-2}$$