Solve 2/4a^2-4a-3-1/4a^2-1

$$\frac { 2 } { 4 a ^ { 2 } - 4 a - 3 } - \frac { 1 } { 4 a ^ { 2 } - 1 }$$

Evaluate

$$\frac{1}{\left(2a-3\right)\left(2a-1\right)}$$

Short Solution Steps

Factor 4a^{2}-4a-3. Factor 4a^{2}-1.

$$\frac{2}{\left(2a-3\right)\left(2a+1\right)}-\frac{1}{\left(2a-1\right)\left(2a+1\right)}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(2a-3\right)\left(2a+1\right) and \left(2a-1\right)\left(2a+1\right) is \left(2a-3\right)\left(2a-1\right)\left(2a+1\right). Multiply \frac{2}{\left(2a-3\right)\left(2a+1\right)} times \frac{2a-1}{2a-1}. Multiply \frac{1}{\left(2a-1\right)\left(2a+1\right)} times \frac{2a-3}{2a-3}.

$$\frac{2\left(2a-1\right)}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)}-\frac{2a-3}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)}$$

Since \frac{2\left(2a-1\right)}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)} and \frac{2a-3}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2\left(2a-1\right)-\left(2a-3\right)}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)}$$

Do the multiplications in 2\left(2a-1\right)-\left(2a-3\right).

$$\frac{4a-2-2a+3}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)}$$

Combine like terms in 4a-2-2a+3.

$$\frac{2a+1}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)}$$

Cancel out 2a+1 in both numerator and denominator.

$$\frac{1}{\left(2a-3\right)\left(2a-1\right)}$$

Expand \left(2a-3\right)\left(2a-1\right).

$$\frac{1}{4a^{2}-8a+3}$$

Differentiate w.r.t. a

$$\frac{8\left(1-a\right)}{\left(\left(2a-3\right)\left(2a-1\right)\right)^{2}}$$

Steps Using Chain Rule

Factor 4a^{2}-4a-3. Factor 4a^{2}-1.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{2}{\left(2a-3\right)\left(2a+1\right)}-\frac{1}{\left(2a-1\right)\left(2a+1\right)})$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of \left(2a-3\right)\left(2a+1\right) and \left(2a-1\right)\left(2a+1\right) is \left(2a-3\right)\left(2a-1\right)\left(2a+1\right). Multiply \frac{2}{\left(2a-3\right)\left(2a+1\right)} times \frac{2a-1}{2a-1}. Multiply \frac{1}{\left(2a-1\right)\left(2a+1\right)} times \frac{2a-3}{2a-3}.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{2\left(2a-1\right)}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)}-\frac{2a-3}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)})$$

Since \frac{2\left(2a-1\right)}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)} and \frac{2a-3}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{2\left(2a-1\right)-\left(2a-3\right)}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)})$$

Do the multiplications in 2\left(2a-1\right)-\left(2a-3\right).

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{4a-2-2a+3}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)})$$

Combine like terms in 4a-2-2a+3.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{2a+1}{\left(2a-3\right)\left(2a-1\right)\left(2a+1\right)})$$

Cancel out 2a+1 in both numerator and denominator.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{1}{\left(2a-3\right)\left(2a-1\right)})$$

Use the distributive property to multiply 2a-3 by 2a-1 and combine like terms.

$$\frac{\mathrm{d}}{\mathrm{d}a}(\frac{1}{4a^{2}-8a+3})$$

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).

$$-\left(4a^{2}-8a^{1}+3\right)^{-1-1}\frac{\mathrm{d}}{\mathrm{d}a}(4a^{2}-8a^{1}+3)$$

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}.

$$-\left(4a^{2}-8a^{1}+3\right)^{-2}\left(2\times 4a^{2-1}-8a^{1-1}\right)$$

Simplify.

$$\left(4a^{2}-8a^{1}+3\right)^{-2}\left(-8a^{1}+8a^{0}\right)$$

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

$$\left(4a^{2}-8a+3\right)^{-2}\left(-8a+8a^{0}\right)$$

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

$$\left(4a^{2}-8a+3\right)^{-2}\left(-8a+8\times 1\right)$$

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

$$\left(4a^{2}-8a+3\right)^{-2}\left(-8a+8\right)$$