### Video Transcript

Use logarithmic differentiation to determine the derivative of the function π¦ equals negative three π₯ to the power of two π₯.

So the first thing we can do if we want to find the first derivative of our function π¦ equals negative three π₯ to the power of two π₯ is bring negative three, our constant value, outside of the differentiation. Because itβs going to factor it and it makes life easier. So we can say that the derivative of our function is equal to negative three multiplied by the derivative of π₯ to the power of two π₯. And thatβs with respect to π₯.

So in order to differentiate our π₯ to the power of two π₯, what weβre gonna use is an adaptation of our power rule. And that is that if weβve got a function raised to the power of a function, then the derivative of it is equal to that function raised to the power of a function. So in our case here, π of π₯ raised to the power of π of π₯ multiplied by the derivative of ln or the natural logarithm of π of π₯ multiplied by π of π₯.

So therefore, if we apply this, weβre gonna have our derivative is equal to negative three multiplied by π₯ power of two π₯ multiplied by the derivative of ln π₯ multiplied by two π₯. So now if we simplify this, we can say that the derivative is equal to negative three π₯ to the power of two π₯ multiplied by the derivative of two π₯ ln π₯.

So now to enable us to differentiate two π₯ ln π₯, what weβre gonna have to use is the product rule. And what the product rule tells us is that if weβve got two functions multiplied together, so π of π₯ multiplied by π of π₯, then if we let π’ equal π of π₯ and π£ equals π of π₯, then the derivative of π’π£ is equal to π’ dπ£ dπ₯ plus π£ dπ’ dπ₯, which is π’ multiplied by the derivative of π£ plus π£ multiplied by the derivative of π’.

Okay, great, we can now apply that to our scenario. So what we can start with is our π’ being equal to two π₯. So therefore, dπ’ dπ₯ is gonna be equal to two. And thatβs because if we want to differentiate two π₯, what we do is we multiply the exponent by the coefficient, so one by two, which gives us two. And then we reduce the exponent by one, so from one to zero. So we just get two multiplied by π₯ to the power of zero, which is just two multiplied by one, which gives us two. And now if we say that π£ is equal to ln π₯ or the natural logarithm of π₯, then weβre gonna have to differentiate this to find dπ£ dπ₯.

So now what we do is apply the natural logarithm differentiation rule. And when we do that, we get dπ£ dπ₯ is equal to one over π₯. And thatβs because we know that the derivative of ln π₯ is equal to one over π₯. So now if we bring that all back together and apply the product rule. So now if we do that, weβre gonna have derivative is equal to negative three π₯ to the power of two π₯ multiplied by. And weβve got π’ dπ£ dπ₯, which is gonna give us two π₯ over π₯. And thatβs because π’ was two π₯ and dπ£ dπ₯ was one over π₯. Then plus π£ dπ’ dπ₯, which would give us ln π₯ multiplied by two. So I brought them together, which gives us two ln π₯.

So now if we look to simplify, the first thing we can do is divide the first term inside the parentheses by π₯ on the numerator and denominator because itβs a common term. So weβre left with two. So weβre now at the stage where the derivative is equal to negative three π₯ to the power of two π₯ multiplied by two plus two ln π₯.

Thereβs one more stage to go with simplifying because we can see that two is a common factor in both terms inside the parentheses. So if we bring that outside the parentheses and we multiply it by negative three π₯ to the power of two π₯, weβre gonna get negative six π₯ to the power of two π₯ multiplied by one plus ln π₯. So therefore, we can say that weβve used logarithmic differentiation to determine the derivative of π¦ equals negative three π₯ to the power of two π₯. And it is, as weβve already stated, negative six π₯ to the power of two π₯ multiplied by one plus ln π₯.