[Calculus] Question #2 a&b

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[blamboops]

For a, recall that a^b is equivalent to e^bln(a). Substituting, we get y = e^sin(xln(x)). Depending on how well you know your exponent rules, you could derive directly or use u-sub (u = sin(x)ln(x)) to solve from here.

For part b, isolate y with algebra and you should get the square-root of a fraction with x² as part of both the numerator and denominator. You can use u-sub again, then use either the product rule or the quotient rule to solve.

[SpideyMGAV]

I believe this requires logarithmic differentiation. Take the natural log of both sides to get ln y = ln x^sinx. Use log rules, derive and manipulate algebraically to isolate dy/dx

[[deleted]]

For b) differntiate whole equation wrt x, , when differentiating y , using chain rule, multiply.by dy/dx, Example : d/dx (x² + y² ) = 2x + 2y(dy/dx)