How do you differentiate f(x)=tan5x^3?

1 Answer
Dec 5, 2016

f'(x)=((v(u(x)))'=15x^2sec^2(5x^3)

Explanation:

Differentiating f(x) is determined by applying the chain rule .
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Let " "u(x)=5x^3" "and " "v(x)=tanx
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Then,
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f(x)=v(u(x))
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Applying the chain rule for differentiation.
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f'(x)=((v(u(x)))'=v'(u(x))xxu'(x)
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u'(x)=15x^2
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v'(x)=sec^2x" "then" "v'(u(x))=sec^2(5x^3)
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Therefore,
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f'(x)=((v(u(x)))'=v'(u(x))xxu'(x)
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f'(x)=((v(u(x)))'=sec^2(5x^3)xx15x^2
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f'(x)=((v(u(x)))'=15x^2sec^2(5x^3)