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ln(x^4) ln(x^4) @Mathematics
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Anyone out there?
what is your question?
4ln(x) ?
Slaabaik is right.
ln(x^4+y^4) differentiate wrt x and y seperately
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(4x^3 + dy/dx 4y^3)/(x^4 + y^4)
differentiate wrt x only; i will figure the other out pls
must I differentiate ln(x^4)? okay. answer is: 4x^3/x^4 = 4/x
@slaaibak no
f=ln(x^4+y^4) differentiate wrt x
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4x^3/(y^4 + x^4)
\[z=\ln (x^{4}+y^{4})\]this can be done in parts, so let:\[w=x^{4}\]then:\[z=\ln(w+y^{4})\]\[dz/dx=dz/dw*dw/dx=\frac{1}{w+y^{4}}*4x^{3}=\frac{4x^{3}}{x^{4}+y^{4}}\]
Thanks guyz
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