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प्रश्न
Differentiate log (1 + x2) with respect to ax.
उत्तर
Let u = log (1 + x2) and v = ax
u = log (1 + x2)
Differentiating both sides w.r.t.x, we get
`"du"/"dx" = 1/(1 + "x"^2) * "d"/"dx" (1 + "x"^2)`
`= 1/(1 + "x"^2) * (0 + "2x")`
∴ `"du"/"dx" = "2x"/(1 + "x"^2)`
v = ax
Differentiating both sides w.r.t.x, we get
`"dv"/"dx" = "a"^"x" * log "a"`
∴ `"du"/"dv" = ("du"/"dx")/("dv"/"dx") = ("2x"/(1 + "x"^2))/("a"^"x" * log "a")`
∴ `"du"/"dv" = "2x"/("a"^"x" * log "a" * (1 + "x"^2))`
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