Find y' with respect to t for y = (at^6
Find $y'$ with respect to $t$ for $y = \dfrac{at^{6} - b^{6}}{\sqrt{a^{2}-b^{2}}}$, where $a$ and $b$ are constants with $a^{2} \gt b^{2}$.
Explanation of Method
Separate the constant factor before differentiating.
Step 1: Separate the constant factor
The independent variable is $t$; every occurrence of $a$ and $b$ is constant. In particular $\sqrt{a^{2}-b^{2}}$ is a constant denominator, so the expression splits as
\[ y = \frac{a}{\sqrt{a^{2}-b^{2}}}\,t^{6} - \frac{b^{6}}{\sqrt{a^{2}-b^{2}}}. \]
The first part is a constant multiple of $t^{6}$; the second part is a pure constant.
Step 2: Differentiate term by term
The constant term contributes nothing, and the power rule gives $\bigl(t^{6}\bigr)' = 6t^{5}$: $y' = \frac{a}{\sqrt{a^{2}-b^{2}}}\cdot 6t^{5} - 0 = \frac{6at^{5}}{\sqrt{a^{2}-b^{2}}}$.
Step 3: Verification
\[a = 1,\ b = 0:\quad y = t^{6} \;\Longrightarrow\; y' = 6t^{5}, \text{ and the formula gives } \frac{6t^{5}}{1} = 6t^{5} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{6at^{5}}{\sqrt{a^{2}-b^{2}}}, \qquad a^{2} > b^{2}\,}\]
Remark
Remark. The restriction $a^{2} \gt b^{2}$ is needed only for the expression itself to be defined (a positive radicand and a non-zero denominator); it plays no role in the differentiation. The term $b^{6}$ is a constant and disappears entirely, however elaborate it looks.