Question

Differentiate log (1 + x²) with respect to a^x.

Answer 1

Let u = log (1 + x²) and v = a^x

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 = a^x 

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))`