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प्रश्न
Differentiate the following w.r.t.x: `log[sec (e^(x^2))]`
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उत्तर
Let y = `log[sec (e^(x^2))]`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"log[sec (e^(x^2))]`
= `(1)/(sec (e^(x^2))). "d"/"dx"[sec (e^(x^2))]`
= `(1)/(sec (e^(x^2))).sec(e^(x^2))tan(e^(x^2))."d"/"dx"(e^(x^2))`
= `tan(e^(x^2)).e^(x^2)."d"/"dx"(x^2)`
= `tan(e^(x^2)).e^(x^2).2x`
= `2x.e^(x^2)tan(e^(x^2))`.
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