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Naddik [55]
3 years ago
12

The derivative of (x^2 +1)(x^3+ 1) (by definition and basic rule)

Mathematics
1 answer:
dlinn [17]3 years ago
8 0

Answer:

f'(x) = 5x^4 + 2x + 3x^2

Step-by-step explanation:

To find the derivative of this equation we can do two things.

One method is to use the product rule, which states that when f(x) consists of two functions multiplied to each other (meaning f(x) = g(x) * h(x)), the derivative is f'(x) = g'(x)*h(x) + g(x)*h'(x). In simple language, the derivative is found by finding the derivative of x² + 1 and multiplying it with the normal function of x³ + 1, after which you add the product of the nnormal function of x² + 1 and the derivative of x³ + 1.

it might be clearer when I show you:

f(x) = (x^2 + 1)(x^3 + 1)\\f'(x) = (2x)(x^3 + 1) + (x^2 + 1)(3x^2)\\Simplify:\\f'(x) = 2x^4 + 2x + 3x^4 + 3x^2 = 5x^4 + 3x^2 + 2x\\

If you are not familiar with this rule you can first write out the function and then use the basic rule:

f(x) = (x^2 +1)(x^3+1) = x^5 + x^2 + x^3 + 1\\f'(x) = 5x^4 + 2x + 3x^2

If you need any further help please say so in the comments! I hope this helps! If the steps seem complicated, I suggest you could revise expanding brackets (the first step of the second method) and the basic rules of deriving, but feel free to reach out if you struggle afterwards still

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Answer:

Therefore the y-intercept of the function is 4.

Step-by-step explanation:

Intercepts:

The line which intersect on x-axis and y-axis are called intercepts.

y-intercept: The line or function which intersect at y-axis. So when the line intersect at y-axis, X coordinate is zero.

So in the given Function Put x = 0 we will get the y-intercept

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Put x =0

f(0)= 4-5\times 0

f(0)= 4-0=4

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