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G(t)=5Cos^2(pi)(t) using the Chain rule

User Sajeev
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\bf g(t)=5cos^2(\pi t)\iff g(t)=5[cos(\pi t)]^2\\\\ -----------------------------\\\\ \textit{thus, using the chain-rule} \\\\\\ \cfrac{dg}{dt}=5\cdot 2cos(\pi t)^1\cdot \pi \cdot t^0\implies \cfrac{dg}{dt}=10\pi cos(\pi t)

one thing to recall with trigonometric functions is that, they have the exponent next to the function, not comprising the argument as well
but the exponent applies to the whole thing

so say
\bf cos^5\left( (2\pi )/(3)\theta \right)\iff \left[ cos\left( (2\pi )/(3)\theta \right) \right]^5

User Asitmoharna
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