### Video Transcript

Differentiate π§ of π¦ is equal to π΄ divided by five π¦ cubed plus π΅ times π to the power of three π¦.

The question gives us π§ which is a function of π¦, and it asks us to differentiate this. So we need to differentiate π΄ divided by five π¦ cubed plus π΅ times π to the power of three π¦ with respect to π¦. So we want to find the derivative of our function π§ with respect to π¦. We notice itβs the sum of two functions, and we can differentiate each of these two terms separately.

To help us differentiate our first term, by using our laws of exponents, we can rewrite it as π΄ times π¦ to the power of negative three divided by five. We can then evaluate this derivative by using the power rule for differentiation, which tells us, for constants π and π, the derivative of ππ¦ to the πth power with respect to π¦ is equal to π times π times π¦ to the power of π minus one. We multiply by the exponent and then reduce the exponent by one.

So letβs use this to differentiate our first term. This gives us π΄ over five multiplied by our exponent, negative three, multiplied by π¦ to the power of negative three minus one. We now need to evaluate the derivative of our second term. We can do this by recalling, for constants π and π, the derivative of π times π to the power of ππ¦ with respect to π¦ is equal to π times π times π to the power of ππ¦.

We can apply this to differentiate our second term. This gives us π΅ times three times π to the power of three π¦. We can simplify our first term to get negative three π΄ divided by five multiplied by π¦ to the power of negative four. And we can rearrange our second term to get three π΅ times π to the power of three π¦. And we could leave our answer like this. However, weβre going to use our laws of exponents to rewrite π¦ to the power of negative four as one divided by π¦ to the fourth power. And this gives us dπ§ by dπ¦ is equal to negative three π΄ divided by five times π¦ to the fourth power plus three π΅ times π to the power of three π¦.