Practice problem
Find the derivative of y=ecsch−1(πx)y=e^{\text{csch}^{-1}(\pi x)}y=ecsch−1(πx)
y′=−ecsch−1(πx)∣x∣1−π2x2y'=-\frac{e^{\text{csch}^{-1}(\pi x)}}{|x|\sqrt{1-\pi^2x^2}}y′=−∣x∣1−π2x2ecsch−1(πx)
y′=−ecsch−1(πx)x1−π2x2y'=-\frac{e^{\text{csch}^{-1}(\pi x)}}{x\sqrt{1-\pi^2x^2}}y′=−x1−π2x2ecsch−1(πx)
y′=−ecsch−1(πx)∣πx∣1−π2x2y'=-\frac{e^{\text{csch}^{-1}(\pi x)}}{|\pi x|\sqrt{1-\pi^2x^2}}y′=−∣πx∣1−π2x2ecsch−1(πx)
y′=−ecsch−1(πx)πxπ2x2+1y'=-\frac{e^{\text{csch}^{-1}(\pi x)}}{\pi x\sqrt{\pi^2x^2+1}}y′=−πxπ2x2+1ecsch−1(πx)
Type a concept, like “slope” or “photosynthesis”.
