Use the chain rule to calculate the derivative of (3-5x)^3(4-x^5)^3
https://www.khanacademy.org/math/calculus/differential-calculus/chain_rule/v/the-chain-rule
this is simple straight forward
I can get as far as this but i dont know what to do after this step\[9(3x-5)^2*(4-x^5)^3+(3x-5)^3*-15x^4(4-x^5)^2\]
your process is off . . . If \[f(x)=x^n\] then \[f \prime (x)=nx ^{n-1}\]
nevermind sorry I jumped the gun on that.
your question has the term \[(3-5x)^3\] while you appear to differentiating \[(3x-5)^3\] which is correct?
Are you asked to simplify this further?
@jotopia34 isn't his derivative wrong for the question he posted. You can see the problem I think he has in my post before this one.
also, I don't think there is a 9 in the solution. I believe it should be \[3(3-5x)^2(-5)(4-x^5)+3(4-x^5)^2(3-5x)^3\]
yes it is. I always think of it as "derivative of the whole first, derivative of inside first" x second part. plus "derivative of the whole second, inside of second" times the first part."
thats what I mean I talked myself out of fixing his coefficient when I was looking at his derivative but now I think he has a copying error somewhere.
lol. I believe you are right rabbit.
I inserted my answer into my homework assignment and it said it was right. I thought I needed to simplify further, but I guess not.
first you bring down the first cubed and multiply by three in the front then you subtract one from the exponent making it squared and then you multiply by the derivative of the inside making the first terms (in the parenthesis) derivative 3(3-5x)^2(-5) then you simplify by multiplying three and negative five making it -15(3-5x)^2. Now for you second term (in parenthesis) do the same and pring down the three subtract one from the exponent making it squared and multiply by the derivative inside the parenthasis making the final simplified answer for that term -15x^4(4-x^5)^2 put it together and your answer is-15(3-5x)^2[-15x^4(4-x^5)^2] and to further simplify you multiply the -15 and the -15x^4
Join our real-time social learning platform and learn together with your friends!