Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    Find dy/dx of y=(2x+3)(x+5)^2
    Okay, so I'm using product rule and this is what i've done so far:

    (2x+3)(x+5)^2
    2 [x+5]^2 + 2x+3 (2) (x+5) (1)
    =2 [x+5]^2 + 2*2x+3 (x+5)

    What do I do next, show the steps please?
    Offline

    17
    ReputationRep:
    (Original post by Nkhan)
    Find dy/dx of y=(2x+3)(x+5)^2
    Okay, so I'm using product rule and this is what i've done so far:

    (2x+3)(x+5)^2
    2 [x+5]^2 + 2x+3 (2) (x+5) (1)
    =2 [x+5]^2 + 2*2x+3 (x+5)

    What do I do next, show the steps please?
    There are a number of ways you could do this. You could expand the 3 brackets and then it'd just be straight differentiation of a polynomial. You could use the product rule

    \frac{d}{dx}uv = \frac{du}{dx}v + u\frac{dv}{dx}

    Where

    u = 2x+3

    and

    v = (x+5)^2

    Or you could do the product rule of 3 functions.

    \frac{d}{dx}uvw = \frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx}

    Where

    u = 2x+3

    and

    v = w = x+5


    A lot of people seem to forget to apply the chain rule, so the best method to ensure you do it correctly would be either to multiply it all out, or the last method, product rule of 3 functions.

    Could you rewrite what you've got up to? What you have done seems to be correct, although don't forget you're multiplying the 2 from the (x+5)^2 with the entire function, not just the 2x ( you didn't put brackets around it).
    Offline

    0
    ReputationRep:
    2(x+5)^2 + 2(2x+3)(x+5)

    2(x+5)[x+5+2x+3]

    2(x+5)(3x+8)
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Noble.)
    There are a number of ways you could do this. You could expand the 3 brackets and then it'd just be straight differentiation of a polynomial. You could use the product rule

    \frac{d}{dx}uv = \frac{du}{dx}v + u\frac{dv}{dx}

    Where

    u = 2x+3

    and

    v = (x+5)^2

    Or you could do the product rule of 3 functions.

    \frac{d}{dx}uvw = \frac{du}{dx}vw + u\frac{dv}{dx}w + uv\frac{dw}{dx}

    Where

    u = 2x+3

    and

    v = w = x+5


    A lot of people seem to forget to apply the chain rule, so the best method to ensure you do it correctly would be either to multiply it all out, or the last method, product rule of 3 functions.

    Could you rewrite what you've got up to? What you have done seems to be correct, although don't forget you're multiplying the 2 from the (x+5)^2 with the entire function, not just the 2x ( you didn't put brackets around it).
    Okay thanks And yeah the simplifying but is the difficult bit for me, I just dont no when and what to cancel out :|
    But thanks for ur help
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Pheylan)
    2(x+5)^2 + 2(2x+3)(x+5)

    2(x+5)[x+5+2x+3]

    2(x+5)(3x+8)
    How did you cancel out? what bits cancelled out each other? Thats the hardest part for me :|
    Offline

    0
    ReputationRep:
    (Original post by Nkhan)
    How did you cancel out? what bits cancelled out each other? Thats the hardest part for me :|
    factorise 2(x+5) from both terms, this leaves you with x+5 in the first term and 2x+3 in the second. x+5+2x+3 = 3x+8
    Offline

    17
    ReputationRep:
    (Original post by Nkhan)
    Okay thanks And yeah the simplifying but is the difficult bit for me, I just dont no when and what to cancel out :|
    But thanks for ur help
    \frac{dy}{dx} = 2(x+5)^2 + 2(2x+3)(x+5)

    You can take out a factor of 2 and (x+5) from both.

    \frac{dy}{dx} = 2(x+5)[(x+5) + (2x+3)]
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Pheylan)
    factorise 2(x+5) from both terms, this leaves you with x+5 in the first term and 2x+3 in the second. x+5+2x+3 = 3x+8
    Ahh right, thanks so much
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.