What is d/dx [sin^(-1)(9x)]?
d/dx [sin^(-1)(9x)] क्या है?
Show answer and explanation
B. 9 / sqrt(1-81x^2)
ExplanationStep 1: derivative of sin^(-1)u is u'/sqrt(1-u^2). Step 2: here u=9x, so u'=9, answer 9/sqrt(1-81x^2). Step 3: tip: inverse sine derivative has a square-root denominator.
Step 1: derivative of sin^(-1)u = u'/sqrt(1-u^2)। Step 2: u=9x है इसलिए u'=9, answer 9/sqrt(1-81x^2)। Step 3: tip: inverse sine में denominator square root वाला होता है।