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प्रश्न
Differentiate the following w.r.t.x. :
y = `"e"^x tanx + cos x log x - sqrt(x) 5^x`
उत्तर
y = `"e"^x tanx + (cos x) (log x) - (sqrt(x)) 5^x`
Differentiating w.r.t. x, we get
`("d"y)/("d"x) = "d"/("d"x) ("e"^x tan x) + "d"/("d"x) (cos x log x) - "d"/("d"x) [(sqrt(x)) 5^x]`
= `("e"^x "d"/("d"x) tan x + tan x "d"/("d"x) "e"^x) + [cos x "d"/("d"x) (log x) + log x "d"/("d"x) (cos x)] - [sqrt(x)"d"/("d"x )(5^x) + 5^x "d"/("d"x) sqrt(x)]`
= `"e"^x sec^2x + tan x ("e"^x) + cos x (1/x) + logx(- sin x) - [sqrt(x)(5^x log 5) + 5^x 1/(2sqrt(x))]`
= `"e"^x (sec^2x + tan x) + cos x/x - sinx logx - [(2x 5^x log 5 + 5^x)/(2sqrt(x))]`
= `"e"^x (sec^2x + tan x) + cosx/x - sinx logx - 5^x((2x log5 + 1)/(2sqrt(x)))`
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