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## 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 $\sqrt{7}$.