2 years ago

If f(x) = cos(ln(x5)), find f ′(1).

2 answers:

5 0

$f(x)=\cos[\ln(x^5)]\implies \stackrel{chain~rule}{\cfrac{df}{dx}=-\sin[\ln(x^5)]\cdot \cfrac{1}{x^5}\cdot 5x^4} \\\\\\ \left. \cfrac{df}{dx}=-\cfrac{5\sin[\ln(x^5)]}{x} \right|_{x=1}\implies -\cfrac{5\sin[\ln(1^5)]}{1}\implies -\cfrac{5\sin(0)}{1}\implies 0$

Westkost [7]2 years ago

3 0

Answer:

Step-by-step explanation:

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Ierofanga [76]

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3 years ago

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Expand the polynomial and simplify: (4xy− 1 8 y^2)^2

Vinvika [58]

Answer: $4y^{2}(2x-9y)^{2}$

Step-by-step explanation:

We have the following polynomial:

$(4xy-18y^{2})^{2}$

This is a polynomial of the form $(a-b)^{2}=a^{2}-2ab+b^{2}$ . Following this rule to expand it, we have:

$(4xy)^{2}-2(4xy)(18y^{2})+(18y^{2})^{2}$

$16x^{2}y^{2}-144xy^{3}+324y^{4}$

Applying common factor $4y^{2}$ :

$4y^{2}(4x^{2}-36xy+81y^{2})$

Note the polynomial inside the parenthesis is a perfect square trinomial, which can be factored to $(2x-9y)^{2}$ . Hence, the final simplification is: