Computing Derivatives Exercises

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• sin x provides the input to cos (□), therefore sin x is the inside function and cos(□) is the outside function.

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• x³ is the input to ln (□), therefore x³ is the inside function and ln(□) is the outside function.

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• We would first figure out 7x and then find 4 raised to that power, so 7x is the inside function and 4^{□} is the outside function.

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• (3x² + 1) is inside the function (□)⁴, therefore (3x² + 1) is the inside function and (□)⁴ is the outside function.

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• This is like our first example.  is the inside function because we need to compute this first. Then we use this as the input to our outside function, e^{□}.

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• The outside function is

f(□) = (□)⁹⁹

and the inside function is

g(x) = x⁵ + 4.

The derivative of the outside function is

f ' (□) = 99(□)⁹⁸,

and the derivative of the outside function with the inside function as input is

f ' (g(x)) = 99(x⁵ + 4)⁹⁸.

The derivative of the inside function is

g ' (x) = 5x⁴,

so the final answer is

h ' (x) = f ' (g(x)) · g'(x) = 99(x⁵ + 4)⁹⁸ · 5x⁴.

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• For h(x) = sin(ln x),

the outside function is sin (□) and the inside function is ln x.The derivative of the outside function is

cos(□)

and the derivative of the inside function is

Taking the derivative of the outside function evaluated at the inside function, and multiplying by the derivative of the inside function, we find

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• We can think of the function

h(x) = ln(ln x)

as

h(x) = f(g(x))

where both

f(□) = ln(□)

and

g(x) = ln x.

Then

and

therefore

.

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• For

h(x) = e^{sin x}

the outside function is

e^{□}

and the inside function is

sin x.

The derivative of the outside function is also

e^{□},

and using the inside function for input gets us

e^{sin x}.

Then we multiply by the derivative of the inside function, which is cos x, to reach the final answer:

h ' (x) = e^{sin x} · cos x.

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• The outside function is

f(□) = ln (□)

and the inside function is

g(x) = x³.

Then

and

g ' (x) = 3x²,

therefore

We can simplify this to just .

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For both parts of the problem, we'll need to know the outside and inside functions and their derivatives.The outside function is

f(□) = (□)²

and its derivative is

f ' (□) = 2(□).

The inside function is

g(x) = ln x

and its derivative is

• Since we know f, f ', g, and g ', we can see how the incorrect formulas were constructed.

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This is equal to

f(g ' (x)),

or the outside function evaluated at the derivative of the inside function, which is strange.

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• Using the chain rule correctly,