What is the partial fx(x,y,z,w,p,q,r) of cos(16r^2 + 4x^7p^4q^5) + e^(y^2zw)
I got -sin(16r^2 + 4x^7p^4q^5)*28p^4q^5x^6
But for some reason i just think its not right. Anyone know?
2006-10-19 08:01:40 · 5 answers · asked by Wayne Woj 1 in Science & Mathematics ➔ Mathematics
It's
-sin(16r^2 + 4x^7p^4q^5)d/dx[16r^2 + 4x^7p^4q^5] + e^(y^2zw)d/dx[y^2zw] =
-sin(16r^2 + 4x^7p^4q^5)(28x^6p^4q^5)
so yes, your answer is correct.
2006-10-19 08:23:11 · answer #1 · answered by James L 5 · 0⤊ 0⤋
I know answers have already been given, but I thought I'd put my two cents in and confirm your correctness!
∂f/∂x = - sin( 16r^2 + 4(x^7)(p^4)(q^5) ) * 28(x^6)(p^4)(q^5)
2006-10-19 09:07:44 · answer #2 · answered by sft2hrdtco 4 · 0⤊ 1⤋
-sin(16r^2 + 4x^7p^4q5)*28p^4q^5x^6
i think its right
2006-10-19 08:22:32 · answer #3 · answered by craftyboy 2 · 0⤊ 0⤋
You got it right.
2006-10-19 10:01:52 · answer #4 · answered by mathematician 7 · 1⤊ 0⤋
your answer seems correct,
i got exactly the same
2006-10-19 09:27:48 · answer #5 · answered by Anonymous · 0⤊ 1⤋