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lord [1]
4 years ago
8

Derivative of (2x+5)^4(5x+2)^-2

Mathematics
1 answer:
Sveta_85 [38]4 years ago
3 0
Use chain rule

\frac{d}{dx}f(x)g(x)=f'(x)g(x)+g'(x)f(x)

and also power rule
\frac{d}{dx} x^m=mx^{m-1}

so

well, we will say f(x)=(2x+5)^4 and g(x)=(5x+2)^{-2}
f'(x)=4(2x+5)^3(\frac{d}{dy}(2x+5))=4(2x+5)^3(2)=8(2x+5)^3
g'(x)=-2(5x+2)^{-3}(\frac{d}{dx}(5x+2))=-2(5x+2)^{-3}(5)=-10(5x+2)^{-3}


so
\frac{d}{dx}(2x+5)^4(5x+2)^{-2}=8(2x+5)^3(5x+2)^{-2}+(-10(5x+2)^{-3})(2x+5)^4)
simplify yourself
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6 0
4 years ago
4768 x 1551<br> divided by 10.2
balandron [24]
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3 0
4 years ago
Read 2 more answers
How do I solve this?
Verdich [7]

Consider the point P where the gears meet. When the smaller gear rotates clockwise, the larger one will rotate counterclockwise.

Through one rotation of the smaller gear, P will have traveled the circumference of the smaller gear, which is 8\pi in.

At the same time, a point P' on the larger gear traverses the same distance along the larger gear's circumference. This point traces out an arc that is subtended by some angle \theta. The arc is as long as the smaller gear's circumference.

The measure of a circle's interior angle subtended by an arc is proportional to a complete revolution, i.e. an angular displacement of 2\pi radians:

\dfrac{14\pi\,\mathrm{in}}{2\pi\,\mathrm{rad}}=\dfrac{8\pi\,\mathrm{in}}\theta\implies\theta=\dfrac{8\pi}7\,\mathrm{rad}\approx205.7^\circ

For part 2, we apply the same reasoning to the larger gear. In one full rotation of the larger gear, the point P' traverses the circumference 14\pi in, and so does the point P on the smaller gear.

\dfrac{8\pi\,\mathrm{in}}{2\pi\,\mathrm{rad}}=\dfrac{14\pi\,\mathrm{in}}\theta\implies\theta=\dfrac{7\pi}2\,\mathrm{rad}=630^\circ

A full rotation is 360 degrees, so the smaller gear would have rotated \dfrac{630^\circ}{360^\circ}=1.75 times.

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3 years ago
What is 4 3/4 + -5/8
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I think is 12 1/7 I'm not sure it's correct but I think it is
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