Question

Please find the derivative of $x3^{(2x^2-1)}$. Show all of your work and explain as much as possible.

Answer 1

Hello! :)

$\large\boxed{\frac{dy}{dx} = 3^{(2x^{2} - 1)} (4x^{2}ln3 + 1)}$

Use the Product Rule to solve for the derivative:

$\frac{dy}{dx}= f(x) * g'(x) + g(x) * f'(x)$

We can divide the expression into two different functions:

$f(x) = x\\\\g(x) = 3^{2x^{2} -1}$

Use the Product Rule equation and the following rules to solve for the derivative:

Derivative of an exponential function with base other than e:

$\frac{dy}{dx} a^{u} = a^{u} * u' * lna$

Power rule:

$\frac{dy}{dx} x^{n} = nx^{(n-1)}$

Find the derivatives using the power rule and derivative of an exponential function. Plug these into the product rule equation:

$\frac{dy}{dx} = x(3^{(2x^{2} -1)} * 4x* ln3) + (1)3^{(2x^{2} - 1)}$

Simplify by factoring out a common factor:  $3^{(2x^{2} - 1)}$

$\frac{dy}{dx} = 3^{(2x^{2} - 1)}( x* 4x * ln3) + (1)$

Simplify further:

$\frac{dy}{dx} = 3^{(2x^{2} - 1)} (4x^{2}ln3 + 1)$

This is the final derivative.

Answer 2

Answer:

$3^{2x^2-1}+4ln(3)*3^{2x^2-1}x^2$

Step-by-step explanation:

$x3^{(2x^2-1)}$

~Apply product rules

$\frac{d}{dx}(x)3^{2x^2-1}+\frac{d}{dx}(3x^{2x^2-1})x\\\frac{d}{dx}=1\\(3^{2x^2-1})=4ln(3)*3x^{2x^2-1}x\\1*3^{2x^2-1}+4ln(3)*3^{2x^2-1}xx$

~Simplify

$3^{2x^2-1}+4ln(3)*3^{2x^2-1}x^2$

Best of Luck!