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The 2nd line in your attempt is wrong. If you had posted the work itself rather than an image of your work, I could show you exactly where you went wrong. That's why we insist that you post your work directly here, not just post an image.carl binney said:Homework Statement
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hi could some body please help me factorise this please ? any chance of a few stages would be much appreciated
Homework Equations
The Attempt at a Solution
my attempt , but my solutions say otherwise ?
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There really is no problem statement here, just a mathematical expression which looks suspiciously like it's the the result of a botched attempt at applying the product rule, and perhaps the chain rule.carl binney said:Homework Statement
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Do things a step at a time. Differentiation (taking derivatives) is a very methodical process.carl binney said:hi sorry about the mistakes, the question is
differentiate , y = (2x-5)^3 (x+3)^6
I can get up to where you have to factor the equation, but am unsure what to do from there , any help would be much appreciated ??
carl binney said:differentiate , y = (2x-5)^3 (x+3)^6
You have a mistake in the line above. The exponent on 2x - 5 in the 2nd term should be 3, not 1.carl binney said:= (x+3)^6 * (3(2x-5)^2*2) + (2x-5)^3 * (6(x+3)^5)
= (x+3)^6 *(6(2x-5)^2) + (2x-5)^3 * (6(x+3)^5)
= 6(x+3)^6 (2x-5)^2+ 6(2x-5) (x+3)^5
carl binney said:factoring this part is where I get stuck ?
carl binney said:hi sorry about the mistakes, the question is
differentiate , y = (2x-5)^3 (x+3)^6
I can get up to where you have to factor the equation, but am unsure what to do from there , any help would be much appreciated ??
carl binney said:= (x+3)^6 * (3(2x-5)^2*2) + (2x-5)^3 * (6(x+3)^5)
= (x+3)^6 *(6(2x-5)^2) + (2x-5)^3 * (6(x+3)^5)
= 6(x+3)^6 (2x-5)^2+ 6(2x-5) (x+3)^5
factoring this part is where I get stuck ?
What the OP is stuck on is not the actual differentiation, but the factoring, so I did not move this thread.epenguin said:This is calculus, not precalculus.![]()
The chain rule product rule is a mathematical rule used in calculus to find the derivative of a function that is the product of two other functions. It is used when one function is nested inside another function.
To apply the chain rule product rule, you first find the derivatives of each individual function. Then, you multiply the two derivatives together and add the product of the original two functions multiplied by the derivative of the inner function.
The chain rule product rule is important because it allows us to find the derivative of more complex functions by breaking them down into simpler functions. This is useful in many areas of science and engineering, where functions can be very complicated.
Yes, the chain rule product rule can be applied to any number of functions that are multiplied together. It is important to follow the same steps of finding the derivatives of each individual function and then multiplying them together.
The chain rule product rule has many real-life applications in fields such as physics, engineering, economics, and biology. For example, it can be used to find the rate of change in chemical reactions, the optimal production level in economics, and the growth rate of populations in biology.