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प्रश्न
Find `"dy"/"dx"` if, y = `"a"^((1 + log "x"))`
उत्तर
y = `"a"^((1 + log "x"))`
Differentiating both sides w.r.t.x, we get
`"dy"/"dx" = "d"/"dx" "a"^((1 + log "x"))`
`= "a"^((1 + log "x")) * log "a" * "d"/"dx" (1 + log "x")`
`= "a"^((1 + log "x")) * log "a" * (0 + 1/"x")`
∴ `"dy"/"dx" = "a"^((1 + log "x")) * log "a" * 1/"x"`
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