Question

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Answer 1

Answer:

$f^{'}(x) = [2^{x^{2}}.sin^{2}2x\sqrt{x^{2}+3}][3x.sin2x + 2(x^{2}+3)(3cos2x+xsin2x.ln2)]$

Step-by-step explanation:

Let $f(x) = (x^{2}+3)^{\frac{3}{2}}.sin^{3}2x.2^{x^{2} }$

By using multiplication rule and chain rule, we get

⇒$f^{'}(x) = [\frac{3}{2}(x^{2}+3)^{\frac{3}{2}-1}.2x].[sin^{3}2x.2^{x^{2}}] + (x^{2}+3)^{\frac{3}{2}}.[(3sin^{3-1}2x.cos2x.2).2^{x^{2}}+sin^{3}2x.2^{x^{2}}ln2.2x]$

or $f^{'}(x) = [3x\sqrt{x^{2}+3}].[sin^{3}2x.2^{x^{2}}] + (x^{2}+3)^{\frac{3}{2}}.[(6sin^{2}2x.cos2x).2^{x^{2}}+2xsin^{3}2x.2^{x^{2}}ln2]$

or $f^{'}(x) = [sin^{2}2x\sqrt{x^{2}+3} ][3x.sin2x.2^{x^{2}} + (x^{2}+3)(6cos2x.2^{x^{2}}+2xsin2x.2^{x^{2}}ln2)]$

or $f^{'}(x) = [sin^{2}2x\sqrt{x^{2}+3}.2^{x^{2}}][3x.sin2x + (x^{2}+3)(6cos2x+2xsin2x.ln2)]$

or $f^{'}(x) = [sin^{2}2x\sqrt{x^{2}+3}.2^{x^{2}}][3x.sin2x + 2(x^{2}+3)(3cos2x+xsin2x.ln2)]$

or $f^{'}(x) = [2^{x^{2}}.sin^{2}2x\sqrt{x^{2}+3}][3x.sin2x + 2(x^{2}+3)(3cos2x+xsin2x.ln2)]$