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If y= lnx prove that dy/dx=1/x
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by definition.... int(1/x) = lnx
y = ln x e^y = x now differentiate both sides with respect to x
use the definition of f'(x)= lim h-->0 ([f(x+h)- f(h)/h)
so u have f(x)=ln x \( f'(x) = \lim_{h \rightarrow 0}\frac{f(x+h)-f(h)}{h}\)
\( f'(x)=\lim_{h \rightarrow 0}\frac{\ln (x+h)- \ln h} {h}\)
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e^y = x e^y dy/dx = 1
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