C++ Basic 2018 (Batch-2)
✥Applied 📐 Derivative of inverse csc functions. 📉 Y=csc⁻¹(secx) find dy/dx? 🤔 Hint: csc⁻¹ means inverse of csc 💡 @ProgrammingWithFaith2018
✅Explanation
Hint:(csc⁻^1)'=
-f'(x) / f(x)sqrt((f(x)^2+1))
=-(secx)'/secx sqrt((secx)^2+1))
=-(-tanx secx) / secx sqrt(sec^2+1)
=tanx secx / secx tanx
=1
@ProgrammingWithFaith2018
Hint:(csc⁻^1)'=
-f'(x) / f(x)sqrt((f(x)^2+1))
=-(secx)'/secx sqrt((secx)^2+1))
=-(-tanx secx) / secx sqrt(sec^2+1)
=tanx secx / secx tanx
=1
@ProgrammingWithFaith2018
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applied1_bagashaw_maltot-1.pdf
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➡️Applied Math I - Begashaw
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