Find the derivative of d{x^2(log2^x)+secx}

Steps to solve Find the derivative of d{x^2(log2^x)+secx}/dx

=d{x^2(log2^x)+secx}/dx

=d{x^2(log2^x)}/dx+dsecx/dx

=x^2{d(log2^x)/dx}+(log2^x)(dx^2/dx)+(dsecx/dx)

=(x^2/xln2)+(log2^x)(2x)+secx.tanx

=(x/ln2)+(log2^x)(2x)+secx.tanx

Hence the derivative of d{x^2(log2^x)+secx} is (x/ln2)+(log2^x)(2x)+secx.tanx

Detail Information:-

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