Computing Derivatives Exercises

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•  f'(x) = 1

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• g ' (x) = 1

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• f ' (x) × g ' (x) = 1 × 1 = 1

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• The function (f × g)(x) is equal to

(x + 1)(x) = x² + x.

The derivative of this function is

(f × g)'(x) = 2x + 1.

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1. No. In this example, the derivative of the product of the functions f and g is NOT equal to the product of their derivatives.

Contrasting addition and subtraction rules to multiplication rules,

(f + g)' = f ' + g '

and

(f – g) ' = f ' – g ',

but

(f × g)' ≠ f ' × g '.

What do we do? We need another rule. This one is called the Product Rule, and states that

(fg)' = f ' g + fg '.

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• The function is now called g instead of f, therefore we need to be careful to use the correct letter.

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• First, we rewrite the function.

The quantity  is a constant, therefore

We use the product rule:

And finally we put everything back together:

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• Since 5 is a constant, we don't need to use the product rule here.

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• Now the function is called h.

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• Simplify the function first:

Now the problem is more simplified. h describes a line with slope 12, therefore

h ' (x) = 12.

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• j(x) = ln x + cos x

There is no product here, only addition. And now the function is called j.

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• We use the product rule:

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• We rewrite f (no derivatives yet):

Now we take the derivative of each term:

f ' (x) = 5x⁴ + 6x² – 8x.

Thankfully, we found the same answer we got using the product rule.

The product rule is sometimes useful for checking answers.

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We need to find the derivative of each factor.

(x²)' = 2x

For the other, we need to use the product rule:

Now we need to remember to put everything back together.

Thankfully, this is the same answer we got before.

Multiplying everything out can make it easier to see when two answers are the same.

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One way is to think of the function as

f(x) = (x³sin x)cos x.

In this case,

f ' (x) = (x³sin x)'cos x + (x³sin x)(cos x)'.

We need to use the product rule again to find one of the derivatives we need:

Now we put this back into the product rule:

The other way is to think of the function as

f(x) = (x³)(sin x cos x).

In this case, we'll need to use the product rule to find

(sin x cos x)'

before we can find the derivative of the original function. Here we go:

While cos²x + sin²x = 1, cos²x – sin²x isn't anything in particular; we can't make this any nicer.Now we can use the product rule to find f'.

Thankfully, we found the same answer either way.