Brief calculus finding derivative of function?
Hello people! My question is f(t)=14/t + 12/t^4 + Sqrt 2
Hello people! My question is f(t)=14/t + 12/t^4 + Sqrt 2
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f(t) = 14/t + 12/t⁴ + √2 ← rewrite using negative exponents
f(t) = 14t⁻¹ + 12t⁻⁴ + √2 ← Now, differentiate
f'(t) = -14t⁻² - 48t⁻⁵ + 0
f'(t) = -14/t² - 48/t⁵ ← ANSWER
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f(t) = 14t⁻¹ + 12t⁻⁴ + √2 ← Now, differentiate
f'(t) = -14t⁻² - 48t⁻⁵ + 0
f'(t) = -14/t² - 48/t⁵ ← ANSWER
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f(t) = 14/t + 12/t^4 + √2
f(t) = ( 14t^-1 ) + ( 12t^-4 ) + √2
f '(t) = ( -14t^-2 ) + ( -48t^-5 ) + 0
f '(t) = ( -14/t^2 ) + ( -48/t^5 )
Hope I helped and good luck :)
f(t) = ( 14t^-1 ) + ( 12t^-4 ) + √2
f '(t) = ( -14t^-2 ) + ( -48t^-5 ) + 0
f '(t) = ( -14/t^2 ) + ( -48/t^5 )
Hope I helped and good luck :)
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Differentiate WRT t i assume
f'(t) = -14t^(-2) - 48t^(-5)
= -14/(t^2) - 48/(t^5)
f'(t) = -14t^(-2) - 48t^(-5)
= -14/(t^2) - 48/(t^5)
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f(t) = 14t^-1 + 12t^-4 + sqrt 2
f ' (t) = -14t^-2 - 48t^-5 + 0
f ' (t) = -14/t^2 - 48/t^5
f ' (t) = -14t^-2 - 48t^-5 + 0
f ' (t) = -14/t^2 - 48/t^5
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where is the question? you have given a function of t only! Is that a question?