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प्रश्न
Select the appropriate hint from the hint basket and fill up the blank spaces in the following paragraph. [Activity]:
"Let f(x) = x2 + 5 and g (x) = ex + 3 then
f[g(x)] = .......... and g[f(x)] =...........
Now f'(x) = .......... and g'(x) = ..........
The derivative of f[g(x)] w. r. t. x in terms of f and g is ..........
Therefore `"d"/"dx"[f["g"(x)]]` = .......... and
`["d"/"dx"[f["g"(x)]]]_(x = 0)` = ..........
The derivative of g[f(x)] w. r. t. x in terms of f and g is
Therefore `"d"/"dx"["g"[f(x)]]` = .......... and
`["d"/"dx"["g"[f(x)]]]_(x = -1)` = .........."
Hint basket : `{f'["g"(x)]·"g"'(x), 2e^(2x) + 6e^x, 8, "g"' [ f (x)]· f'(x),2xe^(x^2+5), − 2e^6,e^(2x) + 6e^x + 14, e^(x^2+5) + 3, 2x, e^x}`
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उत्तर
"Let f(x) = x2 + 5 and g(x) = ex + 3 then
f[g(x)] = (ex + 3)2 + 5
f[g(x)] = (ex)2 + 6ex + 9 + 5
f[g(x)] = e2x + 6ex + 14
g[f(x)] = `bb(e^("x"^2 + 5) + 3)`
Now f'(x) = 2x and g'(x) = ex
The derivative of f[g(x)] w. r. t. x in terms of f and g is f'[g(x)].g'(x).
Therefore `"d"/"dx"["f"["g"("x")]]` = `"d"/"dx"["e"^(2"x") + 6"e"^"x" + 14]`
= `"d"/"dx""e"^(2"x") + 6"d"/"dx""e"^"x" + "d"/"dx"14`
= `"e"^(2"x") "d"/"dx"(2"x") + 6"e"^"x" + 0`
`"d"/"dx""f"["g"("x")]` = `bb(2"e"^(2"x") + 6"e"^"x")`
and `"d"/"dx"["f"["g"(x)]]_("x" = 0) = 2"e"^(2(0)) + 6"e"^(0)`
= `2"e"^0 + 6"e"^(0)`
= 2 × 1 + 6 × 1
`"d"/"dx"["f"["g"(x)]]_(x = 0)` = 8
The derivative of g[f(x)] w. r. t. x in terms of f and g is g'[f(x)].f'(x).
Therefore `"d"/"dx"["g"["f"("x")]] = "d"/"dx"("e"^("x"^2 + 5) + 3)`
= `"d"/"dx" "e"^("x"^2 + 5) + "d"/"dx"(3)`
`"d"/"dx"["g"["f"("x")]] = "e"^("x"^2 + 5) "d"/"dx" "x"^2 + 5 + 0 = "e"^("x"^2 + 5) 2"x" = bb(2"xe"^("x"^2 + 5))`
`"d"/"dx"["g"["f"("x")]]_("x" = -1) = 2"xe"^("x"^2 + 5)`
= `2(-1)"e"^((-1)^2 + 5)`
= `-2"e"^(1 + 5)`
= −2e6
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