i just cant figure how to solve this
(-4x^4* sqrt(x)) + (3/ (x^3*sqrt(x)))
this is what i did
-16x^3* 1/2x^-1/2 + 3 * x^-3 * sqrt(x)
-8x^5/2 + 3 * -3x^-4 * 1/2x^-1/2
-8x^5/2 + -9x^-4 * 1/2x^-1/2
-8x^5/2 + -9/2x^-9/2
I don't know if you understood or were able to follow my answer to your other question that is why I avoided answering this one a few days ago.
Let me answer this now (there are none/not many questions to answer here anyway) for it is weekend here, no work, and plenty of spare time.
If again you fail to acknowledge this answer, whether you understand it or not, then this will be my last answer to your questions.
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Find the 1st derivative of
f(x) = (-4x^4* sqrt(x)) + (3/ (x^3*sqrt(x)))
We can simplify that into
f(x) = -4(x^4)(x^(1/2)) +3(x^-3)(x^(-1/2))
f(x) = -4(x^4.5) +3(x^-3.5)
Then,
f'(x) = -4[4.5*x^3.5] +3[-3.5*x^-4.5]
f'(x) = -18x^3.5 -10.5x^-4.5 --------------answer
Or, in the form of f(x),
f'(x) = -18x^3 *sqrt(x) -10.5/(x^4 *sqrt(x)) ----answer.
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If you don't want to simplify the given f(x), then use
d/dx (u*v) = u*(dv/dx) +v*(du/dx) -----------***
d/dx (u/v) = [v*(du/dx) -u(dv/dx)]/(v^2) -----***
f(x) = (-4x^4* sqrt(x)) + (3/ (x^3*sqrt(x)))
f'(x) = (-4){(x^4)(0.5*x^-0.5) +(x^0.5)(4x^3)} +{(x^3 *x^0.5)*(0) -3[(x^3)*(0.5 x^-0.5) +(x^0.5)*(3x^2)]}/(x^6 *x)
f'(x) = -2(x^4)/(x^0.5) -16(x^0.5)(x^3) -[1.5(x^3)*(x^-0.5) +9(x^2)*(x^0.5)]/(x^7)
f'(x) = -2(x^4)[(x^0.5) /x] -16(x^0.5)(x^3) -1.5/[(x^4)*(x^0.5)] -9(x^0.5)/(x^5)
f'(x) = -2(x^3)(x^0.5) -16(x^0.5)(x^3) -1.5 /[(x^4)*(x^0.5)] -9[x /(x^0.5)]/(x^5)
f'(x) = -18(x^3)(x^0.5) -1.5 /[(x^4)*(x^0.5)] -9[1 /[(x^4)(x^0.5)]]
f'(x) = -18(x^3)(x^0.5) -1.5 /[(x^4)*(x^0.5)] -9 /[(x^4)(x^0.5)]
f'(x) = -18(x^3)(x^0.5) -10.5 /[(x^4)*(x^0.5)]
f'(x) = -18(x^3)sqrt(x) -10.5 /[(x^4)sqrt(x)] -----------same as above.
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Note:
sqrt(x) = x / sqrt(x) ----------***
1/sqrt(x) = sqrt(x) / x ---------***