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प्रश्न
Differentiate the following w.r.t. x:
cos (log x + ex), x > 0
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उत्तर
Let, y = cos (log x + ex)
Differentiating both sides with respect to x,
`dy/dx = d/dx cos (log x + e^x)`
= `-sin (log x + e^x) [d/dx log x + d/dx e^x]`
= `-sin (log x + e^x) [1/x + e^x]`
= `-sin (log x + e^x) ((1 + x e^x)/x)`
= `-1/x (1 + x e^x) sin (log x + e^x)`, x > 0
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