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Nataly [62]
3 years ago
8

Find dy/dx y = 5/8x^3/7 + 8x^4 + 6x - 9

Mathematics
1 answer:
aleksandrvk [35]3 years ago
5 0

Answer:

15/56x^(-4/7) + 32x^3 + 6.

Step-by-step explanation:

Using  the rule for algebraic differentiation:

If  y = ax^n, dy/dx = anx^(n-1).

So :

If  y = 5/8x^3/7 + 8x^4 + 6x - 9

dy/dx = 5/8* 3/7 x^(3/7-1) + 8*4 x^(4-1) + 6x(1-1)

=  15/56x^(-4/7) + 32x^3 + 6.

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Answer:

\dfrac{1}{2\sqrt{x}}

Step-by-step explanation:

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f(x+h) = \sqrt{x+h} = (x+h)^{\frac{1}{2}}

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This can be rewritten as

[x(1+\dfrac{h}{x})]^{\frac{1}{2}}

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From the expansion

(1+x)^n=1+nx+\dfrac{n(n-1)}{2!}+\ldots

Setting x=\dfrac{h}{x} and n=\frac{1}{2},

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Multiplying by x^{\frac{1}{2}},

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x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}=\dfrac{h}{2x^{\frac{1}{2}}}-\dfrac{h^2}{8x^{\frac{3}{2}}}+\ldots

\dfrac{x^{\frac{1}{2}}(1+\dfrac{h}{x})^{\frac{1}{2}}-x^{\frac{1}{2}}}{h}=\dfrac{1}{2x^{\frac{1}{2}}}-\dfrac{h}{8x^{\frac{3}{2}}}+\ldots

The limit of this as h\to 0 is

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8 0
3 years ago
Given cosα = −3/5, 180 < α < 270, and sinβ = 12/13, 90 < β < 180
torisob [31]

I got

-  \frac{6 3}{65}

What we know

cos a=-3/5.

sin b=12/13

Angle A interval are between 180 and 270 or third quadrant

Angle B quadrant is between 90 and 180 or second quadrant.

What we need to find

Cos(b)

Cos(a)

What we are going to apply

Sum and Difference Formulas

Basics Sine and Cosines Identies.

1. Let write out the cos(a-b) formula.

\cos(a - b)  =  \cos(a)  \cos(b)  +  \sin(a)  \sin(b)

2. Use the interval it gave us.

According to the given, Angle B must between in second quadrant.

Since sin is opposite/hypotenuse and we are given a sin b=12/13. We. are going to set up an equation using the pythagorean theorem.

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( - 3) {}^{2}  +  {x}^{2}  =  {5}^{2}

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{x}^{2}  = 16

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sin a =-4/5.

Now use cosine difference formula

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Hope this helps

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