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प्रश्न
Differentiate the following w.r.t.x. :
y = `sinx logx + "e"^x cos x - "e"^x sqrt(x)`
उत्तर
Let y = `sinx logx + "e"^x cos x - "e"^x sqrt(x)`
∴ `("d"y)/("d"x) = "d"/("d"x) [sinx * log x + "e"^x cos x - "e"^x sqrt(x)]`
= `"d"/("d"x) [sinx * logx] + "d"/("d"x) ["e"^x cos x] - "d"/("d"x) ["e"^x sqrt(x)]`
= `(sin x) "d"/("d"x) (log x) + (log x) "d"/("d"x) (sin x) + "e"^x "d"/("d"x) (cos x) + cos x "d"/("d"x) ("e"^x) - ["e"^x "d"/("d"x) (sqrt(x)) + sqrt(x) "d"/("d"x) ("e"^x)]`
= `(sin x) (1/x) + (log x)(cos x) + "e"^x (- sin x) + cos x ("e"^x) - ["e"^x (1/(2sqrt(x))) + sqrt(x)("e"^x)]`
= `sinx/x + (cos x)(log x) - "e"^x sin x + "e"^x cos x - "e"^x/(2sqrt(x)) - "e"^xsqrt(x)`
= `sinx/x + (cos x)(log x) + "e"^x(- sinx + cos x - 1/(2sqrt(x)) - sqrt(x))`
= `sinx/x + (cos x)(log x) + "e"^x [- sin x + cos x - ((1 + 2x)/(2sqrt(x)))]`
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