# differentiate the equationy=tan[ln(ax+b)]

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### 1 Answer

y is a composite function in the form `y=f(g(x))` with `f(X)=tan(X)` and `g(x)=ln(ax+b)` .

`f'(X)=1+tan^2X`

`g'(x)=a/(ax+b)`

then` y'=g'(x)*f'(g(x))=a/(ax+b)*(1+tan^2(ln(ax+b)))`

**The derivative is**

`y'(x)=a(1+tan^2(ln(ax+b)))/(ax+b)`

**Remark**. There is a second expression for the derivative of tan X:

`tan'X=1/cos^2X`

It is the same function but written with a different expression. Using this expression in the derivative of the composite fonction we find a different expression of the **same** derivative:

`y'(x)=a/((ax+b)cos^2(ln(ax+b)))`

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