Question

Find the derivative of the function 2^(9-x^2)?

Answer 1

`dy/dx = ln(2)(-2x)2^(9-x^2)`

Explanation:

Let `y = 2^(9-x^2)`

Use the natural logarithm on both sides:

`ln(y) = ln(2^(9-x^2))`

NOTE: One must use the natural logarithm, because its derivative is well known.

A property of logarithms is that they allow one to move the exponent within the argument to the outside as a factor:

`ln(y) = ln(2)(9-x^2)`

Differentiate both sides:

`(d(ln(y)))/dx = (d(ln(2)(9-x^2)))/dx`

Differentiation is linear, therefore, we can move the constant `ln(2)` to the outside:

`(d(ln(y)))/dx = ln(2)(d(9-x^2))/dx`

We must use the chain rule on the left:

`(d(ln(y)))/dy dy/dx = ln(2)(d(9-x^2))/dx`

The differentiation is now trivial:

`1/y dy/dx = ln(2)(-2x)`

Multiply both sides by y:

`dy/dx = ln(2)(-2x)y`

Substitute `y = 2^(9-x^2)`:

`dy/dx = ln(2)(-2x)2^(9-x^2)`