Math, asked by likhithrao01, 8 months ago

the derivative of log[sin^-1x^2] wrt x is
pls beg u
answer fast​

Answers

Answered by sreeh123flyback
0

Step-by-step explanation:

d(log[sin^-1(x²)]/dx

using chain rule

dy/dx(f(g(h(x))))

=f'(g(h(x)))×g'(h(x))×h'(x)×d(x)/dx

so using this rule we get

dy/dx of log x=1/x

so

dy/dx=1/sin^-1(x²) × 1/root(1-((x²)²) × 2x

dy/dx=2x/root(1-x⁴)sin^-1(x²) is the answer

please read it

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