The internet has gained a reputation as the one place where you can find anything and help with math problems is no exception. You will find hundreds of math problem solvers with steps to help you out with every kind of problem. Use online calculators for a quick check or go through online tutorials for a more in depth understanding of the concepts. With free online math solvers step by step solutions help students understand the method clearly. You can also try posting your math problems online and have it solved by live tutors in no time at all.

## Math Problem Solver with Steps Free

### Solved Examples

**Question 1:**Differentiate y = (2x

^{2}+ 6)

^{3}

**Solution:**

Given y = (2x

Differentiate y with respect to x

$\frac{dy}{dx}$ = $\frac{d}{dx}$ (2x

$\frac{dy}{dx}$ = 3(2x

(by using chain rule)

= 3(2x

= 12x(2x

= 24x

=> $\frac{dy}{dx}$ = 24x

^{2}+ 6)^{3}Differentiate y with respect to x

$\frac{dy}{dx}$ = $\frac{d}{dx}$ (2x

^{2}+ 6)^{3}$\frac{dy}{dx}$ = 3(2x

^{2}+ 6) $\frac{d}{dx}$(2x^{2}+ 6)(by using chain rule)

= 3(2x

^{2}+ 6)(4x)= 12x(2x

^{2}+ 6)= 24x

^{3}+ 72x=> $\frac{dy}{dx}$ = 24x

^{3}+ 72x**Question 2:**Solve $\sqrt{4 - 7x} = \sqrt{2}x$

**Solution:**

Given $\sqrt{4 - 7x} = \sqrt{2}x$

Squaring both side to remove the radical

=> $(\sqrt{4 - 7x})^2 = ( \sqrt{2}x)^2$

=> 4 - 7x = 2x

which is quadratic equation

Solve the quadratic equation

2x

=> 2x

=> 2x(x + 4) - (x + 4) = 0

=> (2x - 1)(x + 4) = 0

either 2x - 1 = 0 or x + 4 = 0

if 2x - 1 = 0

=> 2x = 1

=> x = $\frac{1}{2}$

if x + 4 = 0

=> x = - 4

Step 1:Step 1:

Squaring both side to remove the radical

=> $(\sqrt{4 - 7x})^2 = ( \sqrt{2}x)^2$

=> 4 - 7x = 2x

^{2}=> 2x^{2}+ 7x - 4 = 0which is quadratic equation

Step 2:Step 2:

Solve the quadratic equation

2x

^{2}+ 7x - 4 = 0=> 2x

^{2}+ 8x - x - 4 = 0=> 2x(x + 4) - (x + 4) = 0

=> (2x - 1)(x + 4) = 0

Step 3:Step 3:

either 2x - 1 = 0 or x + 4 = 0

if 2x - 1 = 0

=> 2x = 1

=> x = $\frac{1}{2}$

if x + 4 = 0

=> x = - 4

**Answer:**x = - 4, $\frac{1}{2}$**Question 3:**Find all real solution to the rational equation.

$\frac{2}{x - 1} - \frac{1}{x + 2}$ = 1

**Solution:**

Given, rational equation is $\frac{2}{x - 1} - \frac{1}{x + 2}$ = 1

The LCM of the denominators of the rational expressions is

LCM = (x - 1)(x + 2)

Now multiply both sides of the equations by the LCM and simplify

=> $\frac{2 (x - 1)(x + 2)}{x - 1} - \frac{1 (x - 1)(x + 2)}{x + 2}$ = 1 * (x - 1)(x + 2)

=> 2(x + 2) - (x - 1) = x(x + 2) - 1(x + 2)

=> 2x + 4 - x + 1 = x

=> x + 5 = x

=> x

=> x

=> x

=> x = $\pm\sqrt{7}$

The solutions are, x = $ - \sqrt{7}$ and $\sqrt{7}$.

Step 1:Step 1:

The LCM of the denominators of the rational expressions is

LCM = (x - 1)(x + 2)

**Step 2:**Now multiply both sides of the equations by the LCM and simplify

=> $\frac{2 (x - 1)(x + 2)}{x - 1} - \frac{1 (x - 1)(x + 2)}{x + 2}$ = 1 * (x - 1)(x + 2)

=> 2(x + 2) - (x - 1) = x(x + 2) - 1(x + 2)

=> 2x + 4 - x + 1 = x

^{2}+ 2x - x - 2=> x + 5 = x

^{2}+ x - 2=> x

^{2}+ x - 2 - x - 5 = 0=> x

^{2}- 7 = 0=> x

^{2}= 7=> x = $\pm\sqrt{7}$

The solutions are, x = $ - \sqrt{7}$ and $\sqrt{7}$.