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y=f(a^x) and f(sinx)=lnx find dy/dx
Solution) dy/dx = (a^x)(lnx)f''(a^x), .........(1)but f(sinx) = lnx implies f(x) = ln(arcsinx)hence f''(x) = (1/arcsinx) (1/ ( ( 1-x^2 ) ^ ( !/2 ) ) implies f''(a^x) = (1/arcsin(a^x)) (1/ ((1-a ^ (2x)) ^ (1/2))) ............(2)hence from ...(1) &.....(2) the solution is obtained but it should br noted that the given solution exist only when x belongs to (0,1].
3x^2+19x-14=0
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