The picture is not clear. let me assume
y = (x^4)ln(x^3)
product rule :
d f(x)g(x) = f(x) dg(x) + g(x) df(x)
dy/dx = (x^4)d[ln(x^3)/dx] + d[(x^4)/dx] ln(x^3)
= (x^4)d[ln(x^3)/dx] + 4(x^3) ln(x^3)
look at d[ln(x^3)/dx]
d[ln(x^3)/dx]
= d[ln(x^3)/dx][d(x^3)/d(x^3)]
= d[ln(x^3)/d(x^3)][d(x^3)/dx]
= [1/(x^3)][3x^2] = 3/x
... chain rule (in detail)
end up with
dy/dx = (x^4)[3/x] + 4(x^3) ln(x^3)
= x^3[3 + 4ln(x^3)]
Answer:
Make sure you convert these numbers into one unit. Convert everything to yard. Just convert the feet and inches in yard since it's the biggest unit, and then add them together. You should get a fraction not a whole number.
The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
Given that,
The function is (-x+3)/ (3x-2)
We have to find f(1) and f'(x).
Take the function expression
f(x)= (-x+3)/ (3x-2)
Taking x as 1 value
f(1)= (-1+3)/(3(1)-2)
f(1)=2/1
f(1)=1
Now, to get f'(x)
With regard to x, we must differentiate.
f(x) is in u/v
We know
u/v=(vu'-uv')/ v² (formula)
f'(x)= ((3x-2)(-1)- (-x+3)(3))/ (3x-2)²
f'(x)= ((-3x+2)-(-3x+9))/ 9x²- 12x+4
f'(x)=(-3x+2+3x-9)/ 9x²- 12x+4
f'(x)=2-9/ (9x²- 12x+4)
f'(x)=-7/ (9x²- 12x+4)
Therefore, The function is (-x+3)/ (3x-2) and we get f(1)=1 and differentiation is f'(x)=-7/ (9x²- 12x+4).
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L=2W+3, A=LW, using L from the first in the second gives you:
A=(2W+3)W
A=2W^2+3W, and we are told A=90 so
2W^2+3W=90
2W^2+3W-90=0
2W^2-12W+15W-90=0
2W(W-6)+15(W-6)=0
(2W+15)(W-6)=0, since W>0 for all real possibilities,
W=6ft, and since L=2W+3
L=15ft
So the pool will be 6 ft wide by 15 ft long.
