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Math Help - Derivatives - Confusion over definition

  1. #1
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    Question Derivatives - Confusion over definition

    Hi,

    I'm still really new to calculus, and was confused about the definition of a derivative.

    It was my understanding that the derivative was the slope or gradient at a given point on a curve on a graph.

    However, when you have a function like y = x^5 the curve is constantly growing steeper. So, the derivative of x^5 would be 5x^4.

    Therefore, 5x^4 is equal to 625x? So, assuming the slope is 625x, where exactly is this slope on the graph because the slope is constantly changing as the curve grows steeper?

    Im obviously getting something confused, if someone can help clear this up I'd really appreciate it.

    Thanks.
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  2. #2
    Super Member Deadstar's Avatar
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    Quote Originally Posted by ashleysmithd View Post
    Therefore, 5x^4 is equal to 625x?
    Why..? It's the x that is to the power 4, not the 5.

    Slope of curve at say, x=2 would be 5 \times 2^4 = 80
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  3. #3
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    Quote Originally Posted by ashleysmithd View Post
    Hi,

    I'm still really new to calculus, and was confused about the definition of a derivative.

    It was my understanding that the derivative was the slope or gradient at a given point on a curve on a graph.

    However, when you have a function like y = x^5 the curve is constantly growing steeper. So, the derivative of x^5 would be 5x^4.

    Therefore, 5x^4 is equal to 625x? So, assuming the slope is 625x, where exactly is this slope on the graph because the slope is constantly changing as the curve grows steeper?

    Im obviously getting something confused, if someone can help clear this up I'd really appreciate it.
    The bit in red is where you are going wrong. You seem to be saying that 5x^4 = 5^4x, and that is not true. It's the x that is raised to the power 4, not the 5.

    It is correct that the derivative of x^5 is 5x^4. So for example, if you take x=2, that corresponds to the point (2,32) on the graph of y=x^5. The slope at that point is 5\times 2^4 = 5\times 16 = 80. The same sort of calculation will give you the slope of the graph for any other value of x.

    Edit. Deadstar has just said the exact same thing, even using the same example of taking x=2. Coincidence!
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  4. #4
    Super Member Deadstar's Avatar
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    Quote Originally Posted by Opalg View Post
    The bit in red is where you are going wrong. You seem to be saying that 5x^4 = 5^4x, and that is not true. It's the x that is raised to the power 4, not the 5.

    It is correct that the derivative of x^5 is 5x^4. So for example, if you take x=2, that corresponds to the point (2,32) on the graph of y=x^5. The slope at that point is 5\times 2^4 = 5\times 16 = 80. The same sort of calculation will give you the slope of the graph for any other value of x.

    Edit. Deadstar has just said the exact same thing, even using the same example of taking x=2. Coincidence!
    Haha great minds think alike
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  5. #5
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    Quote Originally Posted by ashleysmithd View Post
    Hi,

    I'm still really new to calculus, and was confused about the definition of a derivative.

    It was my understanding that the derivative was the slope or gradient at a given point on a curve on a graph.

    However, when you have a function like y = x^5 the curve is constantly growing steeper. So, the derivative of x^5 would be 5x^4.

    Therefore, 5x^4 is equal to 625x? So, assuming the slope is 625x, where exactly is this slope on the graph because the slope is constantly changing as the curve grows steeper?
    As has been pointed out, 625x= 5^4 x, not 5x^4. Strictly speaking "slope" is only defined for straight lines. The derivative, at a specific value of x, is the slope of the tangent line to the graph at that point. That's a crucial point in Calculus.

    Im obviously getting something confused, if someone can help clear this up I'd really appreciate it.

    Thanks.
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