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Exercise 5.5 · Q10

Q.Find dydx\frac{dy}{dx} in the following: xcos⁡x+x2+1x2−1x^{\cos x} + \frac{x^2+1}{x^2-1}

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We differentiate a sum of two functions: the first term requires logarithmic differentiation (since both base and exponent depend on xx), and the second term is a rational function handled by the quotient rule. The final derivative is dydx=xcos⁡x(cos⁡xx−sin⁡xlog⁡x)−4x(x2−1)2\frac{dy}{dx} = x^{\cos x} \left( \frac{\cos x}{x} - \sin x \log x \right) - \frac{4x}{(x^2-1)^2}.

The problem asks for dydx\frac{dy}{dx} where

y=xcos⁡x+x2+1x2−1.y = x^{\cos x} + \frac{x^2+1}{x^2-1}.

This is a sum of two very different-looking pieces. The second piece is a straightforward rational function — quotient rule territory. The first piece, xcos⁡xx^{\cos x}, is trickier: the variable xx appears in both the base and the exponent. That’s a classic signal for logarithmic differentiation.

Let’s break it down.


1. Differentiate xcos⁡xx^{\cos x} using logarithmic differentiation

Why can’t we just use the power rule or the exponential rule directly? The power rule (ddxxn=nxn−1\frac{d}{dx} x^n = n x^{n-1}) assumes the exponent is constant. The exponential rule (ddxax=axlog⁡a\frac{d}{dx} a^x = a^x \log a) assumes the base is constant. Here, both change with xx, so neither applies directly.

The trick: take the natural log of both sides, use log properties to bring the exponent down, then differentiate implicitly.

Let u=xcos⁡xu = x^{\cos x}. Then

log⁡u=log⁡(xcos⁡x)=cos⁡x⋅log⁡x.\log u = \log(x^{\cos x}) = \cos x \cdot \log x.

Now differentiate both sides with respect to xx. On the left, by the chain rule:

1u⋅dudx=ddx(cos⁡x⋅log⁡x).\frac{1}{u} \cdot \frac{du}{dx} = \frac{d}{dx} \big( \cos x \cdot \log x \big).

On the right, use the product rule:

ddx(cos⁡x⋅log⁡x)=(−sin⁡x)(log⁡x)+(cos⁡x)(1x).\frac{d}{dx}(\cos x \cdot \log x) = (-\sin x)(\log x) + (\cos x)\left(\frac{1}{x}\right).

So

1u⋅dudx=−sin⁡xlog⁡x+cos⁡xx.\frac{1}{u} \cdot \frac{du}{dx} = -\sin x \log x + \frac{\cos x}{x}.

Multiply through by u=xcos⁡xu = x^{\cos x}:

dudx=xcos⁡x(cos⁡xx−sin⁡xlog⁡x).\frac{du}{dx} = x^{\cos x} \left( \frac{\cos x}{x} - \sin x \log x \right).

Tip

Notice the order: we wrote cos⁡xx−sin⁡xlog⁡x\frac{\cos x}{x} - \sin x \log x, not the other way. It’s just cleaner — but the derivative is the same either way. The key is that both terms appear.


2. Differentiate x2+1x2−1\frac{x^2+1}{x^2-1} using the quotient rule

Let v=x2+1x2−1v = \frac{x^2+1}{x^2-1}. The quotient rule says:

ddx(fg)=f′g−fg′g2.\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2}. …

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